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: 
Standard differentiation gives us:

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

It's well-known that the limit doesn't exist. Hence $f'(0)$ doesn't exist.
Another example. Let
$$
\phi(x) = \begin{cases}
x, & x \ne 0
\\ \; \\
0, &x = 0\end{cases}\tag{1}
$$
In fact, this is exactly the function $\phi(x) = x$, but of course we can write it as (1).

But the point is: you cannot compute $\phi'(0)$ by differentiating the $0$ in (1) to get $0$. That method is just wrong.