Suppose I have some function $f$: $\mathbb{R}_+\rightarrow\mathbb{R}_+$ that is bijective, continuous and strictly increasing. Moreover, it is (at least) twice continuously differentiable everywhere. Since $f$ is bijective, its inverse $f^{-1}$ exists. What conditions need to be satisfied so that $f^{-1}$ is also (at least) twice continuously differentiable everywhere?
From wikipedia's Inverse FT article: when $f$ is a continuously differentiable function with nonzero derivative at the point $a$, then $f^{-1}$ is continuously differentiable. Given that $f$ in my case is strictly increasing $\implies$ $f'>0$ this condition seems satisfied for all $a \in \mathbb{R}_+$.
Counterexamples in comments assume that $f'(x)=0$ at some $a$, but this is ruled out by the assumption that $f'(x)>0$.