Prove or provide a counterexample. Consider $g = f \circ u$ two equivalent continuous curves on an open interval $\mathrm{I}$ with values in $\mathbf{R}^d$ ($d = 1$ is OK). So $u$ is assumed continuous, increasing and in fact differentiable. Suppose $a \in \mathrm{I}$ and $b = u(a).$ Suppose $f$ is not differentiable at $b$ while $g$ and $u$ are both differentiable at $a.$ Does this imply that $g'(a) = 0 \in \mathbf{R}^d$ and $u'(a) = 0 \in \mathbf{R}$?
Here are some partial results.
If $u'(a) \neq 0,$ then $u^{-1}$ (which exists since $u$ is increasing and is continuous since $u$ is) is differentiable at $b$ and $(u^{-1})'(b) = 1/u'(a).$ Then, $f = g \circ u^{-1}$ will be differentiable at $b,$ contradicting the hypothesis. This implies that $u'(a) = 0.$ There remains to prove $g'(a) = 0$ as well.
If $f$ has lateral derivatives at $b,$ then we can apply the chain rule on the left or on the right of the point $a$ (which is valid since $u$ preserves orientation) and conclude that $g'_-(a) = g'_+(a) = 0$ so $g'(a) = 0.$
Therefore, for a counterexample to exist, there has to be a change of parameter $u$ and a curve without lateral limits such that $u'(a) = 0$ and $f \circ u$ has a derivative at $a.$
I think $f(x)=x^\frac{1}{3}$; $u(x)=x^3$; $a=0$ will do as a counter-example: then $g(x)=x$, which does not have $g'(0)=0$.