Curvature of a curve is independent of the parametrization My professor defined the curvature $\kappa$ of a regular curve $\gamma:[a,b]\to\mathbb{R}^n$ at a point $t\in(a,b)$ to be $$k(t) = \frac{\left|\gamma''(t)\right|}{\left|\gamma'(t)\right|^2}$$ but there was not enough time to prove that this definition is independent of the parametrization and I am trying to show that.
Let $\phi:[c,d]\to[a,b]$ be a (smooth) diffeomorphism. For readability reasons, we will denote $\gamma\circ\phi(s)$ by $\gamma$ and $\phi(s)$ by $\phi$. Then 
\begin{array}
ft\kappa^2(t) = \frac{\left|(\gamma\circ\phi)''(t)\right|^2}{\left|(\gamma\circ\phi)'(t)\right|^4} = \frac{|(\gamma'\phi')'|^2}{|\gamma'\phi'|^4} = \frac{|\gamma''\phi'^2+ \gamma'\phi''|^2}{|\gamma'\phi'|^4} &= \frac{\langle\gamma''\phi'^2+ \gamma'\phi'', \gamma''\phi'^2+ \gamma'\phi''\rangle}{|\gamma'\phi'|^4} \\&= \frac{|\gamma''|^2}{|\gamma'|^4} + \phi''\frac{\phi''|\gamma'|+ 2\phi'^2\langle\gamma',\gamma''\rangle}{|\gamma'\phi'|^4}.
\end{array}
By this calculation, if $\gamma$ is reparametrized by arc length, then $\phi'' = 0$, and we have our result. But can I ensure that the expression on the right is 0 for any diffeo $\phi$?

This expression is correct for any constant-speed parametrization, which are all related by affine diffeomorphisms (i.e. $\phi'' = 0$); so your calculation shows invariance within this restricted class.
For formulae in a general coordinate see e.g. wikipedia.
