Let $|f(x)|\leq M$ for an $M>0$ and all $x\in \mathbb{R}$. Let $g(x)=x^2f(x)$ for all $x\in \mathbb{R}$. I need to show that $g$ is differentiable at $0$.
With the definition of differentiability I got $\lim_{h \to 0} \frac{g(0+h)-g(0)}{h}$ = $\lim_{h \to 0} \frac{h^2f(h)}{h}$ = $\lim_{h \to 0} hf(h)$.
I also know that $f(h)$ is bounded by $M$, so as $h$ goes to $0$, $hf(h)$ also goes to 0. (Correct?)
I do however not know if $f(h)$ is continous or differentiable, is this a prerequisite?
This looks fine to me. And no, we don't need any more niceness conditions on $f$. Boundedness is enough. Any function squeezed between $Mx^2$ and $-Mx^2$ (which is what $g(x)$ is), no matter how discontinuous it otherwise is, will be differentiable at $x = 0$. That is what you have proven, after all.
If you really want a nit-pick, you could be a bit more explicit about how $f$ being bounded by $M$ implies that $hf(h)\to 0$. Whether that's actually necessary (and if so, in how much detail), depends on your teacher / professor / TA, since deciding whether you've demonstrated the skill you need to demonstrate on tests and assignments is ultimately up to them.