How can $\frac{1}{\ln(x)}$ be equal to $0$?

$$x \mapsto \frac{1}{\ln(x)}$$ is indeed not defined for $x=0$, but the point is that it can be extended to a continuous function $f : [0,1) \rightarrow \mathbb{R}$ satisfying $f(0)=0$. More precisely, define $$f(x)=\left\{ \begin{array}{ll} \frac{1}{\ln(x)} & \mbox{if } 0<x<1 \\ 0 & \mbox{if } x=0 \end{array} \right. $$

You can prove that $f$, as defined, is a continuous function, and this is the only way to extend your original function to a continuous one at $x=0$.

If you define

$$\cot(x):=\dfrac1{\tan(x)},$$ then the cotangent can never be $0$. But if you define it as

$$\cot(x):=\dfrac{\cos(x)}{\sin(x)},$$ then it is indeed zero for all $x$ such that $\cos(x)=0$. The second is the "best" definition, and

$$\cot(x)\equiv\dfrac1{\tan(x)}\equiv\dfrac{\cos(x)}{\sin(x)}$$ is not completely true.

Now if we look at

$$\frac1{\log(x)},$$

this expression is indeed undefined at $x=0$ because the logarithm is undefined. You can extend the definition in a way that makes the function continuous as follows:

$$\begin{cases}x=0\to\lim_{t\to0}\dfrac1{\log(t)},\\x>0\to\dfrac1{\log(x)}.\end{cases}$$

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Note that a grapher cannot show you the difference between the definitions, because they differ in isolated points, which are "infinitely tiny".

Desmos draws the graph of a function at the points at which it is defined. Indeed, $\frac1{\ln x}$ is undefined at $0$; but, since $\lim_{x\to0^+}\frac1{\ln x}=0$, you see nothing peculiar at $(0,0)$. If you ask Desmos to draw the graph of, say $\frac{\sin(x)}x$ or $\frac{\cos(x)-1}{x^2}$, the same thing will occur.

And the equality $\cot x=\frac1{\tan x}$ is valid only when $\tan(x)\ne0$.