So I'm having a little trouble algebraically showing this is true, the hint is that it is an exercise of the chain rule.
From definition, a parametrised curve $\tilde\gamma : J \rightarrow \mathbb{R}^n$ is a reparametrisation of $\gamma: I \rightarrow \mathbb{R}^n$ if $\exists$ a diffeomorphism $\phi : J \rightarrow I $ such that $\tilde\gamma(\tilde t) = \gamma \circ \phi (\tilde t) \forall \tilde t \in J$.
I feel like the result is obvious, I just don't know quite how to prove it algebraically. I'm sure differentiating both sides would be a good start but past that I'm not sure.
Okay. So the definition of a regular curve, $\mu:I\rightarrow\mathbb{R}^n$ is that $\forall t\in I:\mu'(t)\neq 0$. Assume that $\gamma$ is regular. Now.. a diffeomorphism is regular because it's a smooth bijection defined on an open set with an smooth inverse. With that in mind observe that:
$\tilde{\gamma}'(\tilde{t})=(\gamma(\phi(\tilde{t})))'=\gamma'(\phi(\tilde{t}))\phi'(\tilde{t})$
But since both $\gamma$ and $\phi$ are regular, we know that $\gamma'(\phi(\tilde{t}))\phi'(\tilde{t})\neq 0$ for all $t\in J$.