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: 
It is unfortunate that math gets taught in such a way that people think it's done by applying rules of that sort. It **relies on** many such rules, but they are only a tool.

Actually **understanding** the subject as opposed to applying such rules will remind you that the value of the derivative at a point is the slope of the curve at that point, and that depends on what happens in some interval surrounding that point, not just on the value of the function at that point alone. Thus when you are told that if $f(x) = 5$ then $f'(x) =0,$ it means that if there is some interval such that for all values of $x$ in that interval $f(x) = 5,$ then $f'(x)=0$ for every value of $x$ in the interior of that interval.

To say that $f'(x)=0$ because $f(x) = 5,$ when $f(x) = 5$ at just an isolated point, is nonsense. For example, suppose $f(x) = x^2 - 4.$ Then $f(3) = 3^2-4=5.$ But that doesn't mean $f'(3)=0.$ In fact $f'(x) = 2x,$ so $f'(3)=6.$

Since the usual rules don't apply at $0$ for the function you give us, you need to go back to the definition:
$$
f'(0) = \lim\_{h\to0} \frac{f(0+h) - f(0)} h.
$$
If that limit exists then $f'(0)$ exists; otherwise it doesn't.
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, 
$$\sin \left( \frac{1}{z\_n} \right) = 0~~~~and ~~~~\sin \left( \frac{1}{y\_n} \right) = 1$$