Hey do you think I did alright trying to prove these questions?
a.
Let $f: R->R$, $ \ \ f(x)=\begin{cases} x&x\ne 0\\ 1&x=0 \end{cases}. $
And let $x_{0} = 0$. By the algebra of continuous functions $f$ is differentiable on the deleted neighbour of 0, but $f$ is not continuous at 0, therefore not differentiable at 0.
b.
$f$ is continuous at $x_{0}$, therefore
$\lim _{x\to x_0}\left(f\left(x\right)\right)=f\left(x_0\right)$
Now we check if $f$ is differentiable at $x_0$
$f'(x_0)=\lim _{x\to x_0}\frac{\left(f\left(x\right)-f\left(x_0\right)\right)}{x-x_0}$
We have $\frac{"0"}{0}$, therefore we use L'Hopital
$\lim _{x\to x_0}\frac{\left(f\left(x\right)-f\left(x_0\right)\right)}{x-x_0}=\lim \:_{x\to \:x_0}\left(\:f'\left(x\right)\right)=L$
Therefore $f$ is differentiable at $x_0$.