I was wondering, if there is a function $f$ that is differentiable everywhere in $\mathbb{R}$ (but not continuously differentiable), with a differentiable inverse.
Since it needs to be bijective and it is continuous, it is w.l.o.g. strictly increasing and thus has nonnegative derivative everywhere.
I have tried looking at the standard pathological functions that I know of, but I couldn‘t find any example.
If the restriction to invertible functions with differentiable inverse wasn’t there, then there are some good examples of such pathological functions.
The derivative of an differentiable function with a (differentiable) inverse function is not necessarily continuous.
The function $f: \Bbb R \to \Bbb R$ defined as $$ f(x) = \begin{cases} 4x + x^2 \sin(1/x) & \text{ if } x \ne 0 \, , \\ 0 & \text{ if } x = 0 \,, \end{cases} $$ is differentiable everywhere, with $$ f'(x) = \begin{cases} 4 + 2x \sin(1/x) -\cos(1/x) & \text{ if } x \ne 0 \, , \\ 4 & \text{ if } x = 0 \, , \end{cases} $$ but $f'$ is not continuous at $x=0$.
For all $x \ne 0$ is $|x \sin(1/x)| \le 1$ and $|\cos(1/x)| \le 1$, so that $f'(x) \ge 1$ for all $x \in \Bbb R$. It follows that $f$ is strictly increasing and therefore invertible. The inverse function is differentiable everywhere by the inverse function rule.