Implicit Differentiation Proof

It doesn't make any more sense to "prove implicit differentiation" than it does to "prove numbers," but I assume you're asking why implicit differentiation is valid i.e. preserves the truth of equations.

Implicit differentiation is just an application of the chain and other derivative rules to both sides of an equation, with (in the usual case) $y$ an abridgment of $f(x)$. Observe:

$$\frac{d}{dx}g(f(x),x)=\frac{\partial g}{\partial f}(f(x),x)\cdot\frac{df}{dx}(x)+\frac{\partial g}{\partial x} \tag{1}$$

$$\left(g(y,x)\right)'=\frac{\partial g}{\partial y}y'+\frac{\partial g}{\partial x} \tag{2}$$

The first is how you would take the derivative of a function of both $y=f(x)$ and $x$, but notice the difference between $d/dy$ and $\partial/\partial y$, while the latter equation is the same thing in fewer symbols courtesy of implicit differentiation. You always use ID in contexts where you have a function of both dependent and independent variables, and it's valid because of the above analogy - and it indeed generalizes to arbitrarily many dependent (or independent!) variables. For example, if we understand $u$ and $v$ to be functions of $t$ then we can differentiate

$$u^2+tv^2=1\quad\longrightarrow\quad 2uu'+(1\cdot v^2+t\cdot2vv')=0.$$

Implicit differentiation is simply the use of the chain rule to differentiate a function. Often this makes it possible to differentiate a function that is difficult or impossible to separate into the form $y = f(x)$.

For example, consider the function $y = e^{xy}$. There is no way to separate $y$, so it is impossible to take an 'explicit' derivative. However, we can find an 'implicit' derivative in terms of $x$ and $y$ using the chain rule.

How do we do this? Well, simply take the derivative of both sides: $$ \frac{d}{dx} y = \frac{d}{dx} e^{xy} $$ The left hand side is simply $dy/dx$, or $y'$. The right hand side looks strange to derive, since the equation depends on both y and x. However, a few applications of the chain rule will clear things right up. First, let the 'inside function' be $u = xy$ and the outside function $e^u$. The derivative is, of course, $$ \frac{d}{dx} e^u = \frac{du}{dx} \frac{d}{du} e^u = \frac{du}{dx} e^u $$ Now, using the product rule and then the chain rule, $$ \frac{du}{dx} = \frac{d}{dx} xy = (\frac{d}{dx} x)y + x (\frac{d}{dx} y) = (1)y + x(\frac{dy}{dx} \frac{d}{dy}y) = y + x(\frac{dy}{dx}*1) = y + xy' $$ Finally, our implicit derivative is $$ y' = (y + xy')e^{xy} $$ and we can easily solve for $y'$.

So you see that implicit differentiation is the result of giving up on separating variables to get a derivative in terms of $x$ only, and cleverly using the chain rule to instead get a derivative in terms of $x$ and $y$.