Let $g:\mathbb{R}\rightarrow \mathbb{R}$ be continuous at $0$.
1.) Prove that $f(x)=xg(x)$ is differentiable at $0$ (via proof form).
2.) Briefly explain how/why the continuity of $g$ at $0$ was needed in part (a).
I'm not sure how.
Let $g:\mathbb{R}\rightarrow \mathbb{R}$ be continuous at $0$.
1.) Prove that $f(x)=xg(x)$ is differentiable at $0$ (via proof form).
2.) Briefly explain how/why the continuity of $g$ at $0$ was needed in part (a).
I'm not sure how.
By definition $f$ is differentiable at $0$ if the following limit exists: $\lim_{x→0}\frac{f(x)-f(0)}{x-0}=\lim_{x→0}\frac{xg(x)-0}{x-0}=\lim_{x→0}g(x)$
Because $g$ is continuous at $0$ this limits exist and must equal $g(0)$.
We have $\frac{xg-0}{x-0}=g$. Let the limit approach $0$, you should be able to conclude the limit is $g(0)$. It might be better to try this yourself.
As Jonas Meyer commented, a removable discontinuity is suffice at here.