Why isn't the Euler-Lagrange equation trivial?
Ah, what a tricky mistake you've made there. The problem is that you've simply confused some notions in multivariable calculus. Don't feel bad though-- this is generally very poorly explained. Both steps 1 and 3 above are incorrect. Rest assured, the Euler-Lagrange equation is not trivial.
Let's first take a step back. The Lagrangian for a particle moving in one dimension in an external potential energy $V(q)$ is $$ L(q, \dot q) = \frac{1}{2}m \dot q^2 - V(q). $$ This is how most people write it. However, this is very confusing, because clearly $q$ and $\dot q$ are not independent variables. Once $q$ is specified for all times, $\dot q$ is also specified for all times.
A better way to write the above Lagrangian might be $$ L(a, b) = \frac{1}{2}m b^2 - V(a). $$ Here we've exposed the Lagrangian for what it really is: a function that takes in two numbers and outputs a real number. Likewise, we can clearly see that $$ \frac{\partial L}{\partial a} = -V'(a) \hspace{1cm} \frac{\partial L}{\partial b} = m b. $$ Usually, most people write this as $$ \frac{\partial L}{\partial q} = -V'(q) \hspace{1cm} \frac{\partial L}{\partial \dot q} = m \dot q. $$ However, $q$ and $\dot q$ must be understood as independent variables in order to do this correctly. Just as $a$ and $b$ were independent variables, $q$ and $\dot q$ are too when they're being put into the Lagrangian. In other words, we could put any two numbers into $L$; we just decided to put in $q$ and $\dot q$.
Furthermore, let's look at the total time derivative $\frac{d}{dt}$. How should we understand the following expression? $$ \frac{d}{dt} L(q(t), \dot q(t)) $$ Both $q$ and $\dot q$ are functions of time. Therefore, $L(q(t), \dot q(t))$ depends on time simply because $q(t)$ and $\dot q(t)$ do. Therefore, in order to evaluate the above expression, we need to use the chain rule in multivariable calculus. $$ \frac{d}{dt} L(q(t), \dot q(t)) = \frac{dq}{dt} \frac{\partial L}{\partial a}(q(t), \dot q(t)) + \frac{d \dot q}{dt} \frac{\partial L}{\partial b}(q(t), \dot q(t)) = \dot q(t) \frac{\partial L}{\partial a}(q(t), \dot q(t)) + \ddot q(t) \frac{\partial L}{\partial b}(q(t), \dot q(t)) $$
In the above expression, I once again used $a$ and $b$ in order to make my point clearer. We need to take partial derivatives of $L$ assuming $a$ and $b$ are independent variables. AFTER differentiating, we THEN evaluate $\partial L / \partial a$ and $\partial L / \partial b$ by plugging in $(q, \dot q)$ into the $(a,b)$ slots. This is just like how in single variable calculus, if you have $$ f(x) = x^2 $$ and you want to find $f'(3)$, you first differentiate $f(x)$ while keeping $x$ an unspecified variable, and THEN plug in $x = 3$.
In your first step, the derivatives DON'T commute because $t$ and $q$ are not independent. ($q$ depends on $t$.) Yes, partial derivatives commute, but ONLY if the variables are independent. In your third step, you can't "cancel the dots" because $L$ depends on two inputs. If $L$ only depended on $q$, then yes, you could "cancel the dots" (as this is equivalent to the chain rule in single variable calculus), but it doesn't, so you can't.
EDIT: You can see for yourself that the Euler-Lagrange equation is not identically $0$. If you take the Lagrangian $L(q, \dot q)$ I've written above and plug it into the Euler Lagrange equation, you get $$ m \ddot q(t) + V'(q(t)) = 0. $$ This is not the same as $0 = 0$. It is a condition that a path $q(t)$ would have to satisfy in order to extremize the action. If it was $0 = 0$, then all paths would extremize the action.
EDIT: As Arthur points out, this is also a good time to discuss the difference between $dL / dt$ and $\partial L / \partial t$. If we have a time dependent Lagrangian, $$ L(q, \dot q, t) $$ then $L$ can depend on $t$ explicitly, as opposed to just through $q$ and $\dot q$. So, for example, where as we might have the Lagrangian for a particle in a constant gravitational field $g$ is $$ L(a,b) = \frac{1}{2} mb^2 - m g a $$ if we let allow $L$ to depend on $t$ explicitly, we could have the gravitational field get stronger as time goes on: $$ L(a,b,t) = \frac{1}{2} mb^2 - m ( C t )a. $$ ($C$ is a constant such that $Ct$ has the same units as $g$.)
The quantity $$ \frac{\partial}{\partial t} L(a, b, t) $$ should be understood as differentiating the "$t$-slot" of $L$. In the above example, we would have $$ \frac{\partial}{\partial t} L(a,b,t) = - m C a. $$ The quantity $$ \frac{d}{d t} L(q(t), \dot q(t), t) $$ should be understood as the full time derivative of $L$ due to the fact that $q$ and $\dot q$ also depend on $t$. For the above example, \begin{align*} \frac{d}{d t} L(q(t), \dot q(t), t) &= \dot q(t) \frac{\partial L}{\partial a}(q(t), \dot q(t),t) + \ddot q(t) \frac{\partial L}{\partial b}(q(t), \dot q(t),t) + \frac{\partial L}{\partial t} (q(t), \dot q(t), t) \\ &= (\dot q) (-mC t ) + \ddot q(t) (m \dot q(t)) - mC q(t) \end{align*}
The commutator $$\left[\frac{\partial}{\partial \dot{q}^j},\frac{\mathrm d}{\mathrm d t}\right]~\stackrel{(2)}{=}~\frac{\partial}{\partial q^j}\tag{1}$$ of a velocity derivative $\frac{\partial}{\partial \dot{q}^j}$ with the total time derivative $$\frac{\mathrm d}{\mathrm d t} ~=~\frac{\partial}{\partial t} +\dot{q}^j\frac{\partial}{\partial q^j} +\ddot{q}^j\frac{\partial}{\partial \dot{q}^j} +\dddot{q}^j\frac{\partial}{\partial \ddot{q}^j} +\ldots \tag{2}$$ is not zero. See also e.g. this related Math.SE post & this related Phys.SE post.
The cancellation of dots $$\frac{\partial \dot{L}}{\partial \dot{q}^j}~=~\frac{\partial L}{\partial q^j}\tag{3}$$ works for functions $L(q,t)$ that don't depend on velocities $\dot{q}^k$. But a Lagrangian typically depends on velocities. See also this related Phys.SE post.
Note the following algebraic Poincare lemma: $$L\text{ satisfies the Euler-Lagrange (EL) eqs. identically }$$ $$\quad\Updownarrow\quad\tag{4}$$ $$L\text{ is a total time derivative}$$ (modulo possible topological obstructions). For details, see e.g. this & this Phys.SE posts.
So, in principle one can choose essentially $\it{any}$ Lagrangian $\mathcal{L}$ with sufficiently chosen coordinates (and possibly constraints), and apply variational calculus to it via the Euler-Lagrange equations. The equations of motion that this produces may or may not correspond to an understandable model of reality. There are lots of Lagrangians that don't correspond to reality (seemingly). The Lagrangians that produce physical models have been found usually by guess-and-check and consultation with experiment/observation.
why is this a fundamental law of physics and not a simple triviality of ANY function L on the variables $q$ and $\dot{q}$?
The Euler-Lagrange formalism is not a "fundamental law of physics." Rather, it is a partial differential equation (or a set of them) whose solutions make a particular functional stationary, meaning the solutions obey the principle of extremized action. This mathematical concept was actually generalized in control theory by Pontryagin's maximum principle. The laws of physics are derivable through the Euler-Lagrange method, but the method is not fundamental, similar to how the particular geometry chosen is not fundamental(par. 17) for deriving physical laws. Physicists use math to model reality, so of course we're going to use the things that work! For instance, Einstein derived his field equations heuristically, but Hilbert derived them (around the same time) from the action principle by guessing the correct $\mathcal{L}$. But nowadays, almost everyone that works with general relativity or modified gravity start from $\mathcal{L}$ and use the action principle (except in cosmology they typically start from the metric itself).
It is not entirely surprising that since we are natural creatures which evolved to understand patterns of our environment, the tools we create - especially the abstract ones like math - might have some correspondence with reality. Eugene Wigner wrote a very nice essay about this topic, called "The Unreasonable Effectiveness of Mathematics in the Natural Sciences," in which he argues that it is obvious that math works so well at modeling reality, but it's not at all obvious why this works.
"Why" questions are very difficult to answer, and this one is especially difficult. Some Lagrangians work at producing physical models, and some don't, and maybe the E-L equations work as a filter for figuring that out since it can be used to make testable predictions.
@ AccidentalFourierTransform already clarified your mathematical errors, so I will not.