Question: When we use our standard product and chain rules for when $x$ not equal to $0,$ our derivative function works.

But when $x=0,$ $f'(x)$ does not exist. But why can't we just use standard differentiation rule on $x=0$? Won't that mean $f'(0) = 0$?

Answer: 
That does not even exists.
In fact, $$f'(0) = \lim\_{x \to 0} \frac{f(x) - f(0)}{x-0} = \lim\_{x \to 0} \sin \left( \frac 1x \right)$$
Does not exist . Just consider: $$z\_n =\frac{1}{n\pi}\to 0~~~and~~~y\_n =\frac{1}{\frac{\pi}{2}+2n\pi}\to 0$$
But for all n we have, 
$$\sin \left( \frac{1}{z\_n} \right) = 0~~~~and ~~~~\sin \left( \frac{1}{y\_n} \right) = 1$$
Whenever you have a function with cases for $x=a$ and $x \neq a$, to calculate $f'(a)$ you have to use the definition of the derivative and do it with the limit.

In your case you have a function with cases for $a=0$. Therefore to determinate $f'(a)=f'(0)$ you have to go with the definiton.

$$f'(0) = \lim\_{x \to 0} \frac{f(x) - f(0)}{x-0} = \lim\_{x \to 0} \sin \left( \frac 1x \right)$$

Since that limit does **NOT** exist, you can conclude that $f'(0)$ doesn't exist.