If a function $f:\mathbb R^{n} \rightarrow \mathbb R^m $ is differentiable at a point $a$ can we say that there is a neighbourhood of $a$ such that $f$ is locally Lipschitz? (i.e. there is some constant $M < \infty$ such that for all $x$ in that neighbourhood $||f(x)-f(a)|| < M ||x-a||$).
I can easily prove this is true when the derivative is bounded but then if it is not bounded could I prove there is some small neighbourhood where the derivative is bounded?
Your definition of locally Lipschitz is wrong. At least it is different from wiki.
It should be like this:
There is $M<\infty$ and neighbourhood $U$ of $a$ that for any $x,y \in U$ $\|f(x)-f(y)\|<M\|x-y\|$
Now I can give you counter example. You are looking for this function:
$$ f(x) = x^2 \sin\left(\frac1{x^2}\right) $$
It is everywhere differentiable. But its derivative is not bounded on any neighbourhood of zero.