Show $f$ is not differentiable at $x = 0$. 
Define $f(x) =  \begin{cases} 
      x & x\in \mathbb{Q} \\
      0 & x \in \mathbb{R}\setminus\mathbb{Q}
   \end{cases}$
Is $f$ differentiable at $0$? Is $g(x) = xf(x)$ differentiable at $0$?

Is my reasoning correct? Is there a simpler way to prove $f'(0)$ does not exist? I was typing this up because I was stuck on it but I ended up figuring it out so I thought why not answer my own question.

For $h \neq 0$ you have 
$$\frac{f(h)-f(0)}{h-0}=\begin{cases}
1 & \text{for } h \in \mathbb Q\\
0 & \text{for } h \in \mathbb{R}\setminus\mathbb{Q}
\end {cases}$$
So $\lim\limits_{h \to 0} \frac{f(h)-f(0)}{h-0}$ can’t exist as a limit is unique. Hence $f$ is not differentiable at $0$.
$g$ is differentiable at $0$ and $g^\prime(0)=0$ as for all $x \neq 0$
$$\left\vert \frac{g(x)}{x} \right\vert \le \vert x \vert.$$
