Let $M,N$ be smooth $d$-dimensional Riemannian manifolds.
Suppose $f:M \to N$ is a differentiable isometry ($df_p$ is an isometry at every $p \in M$). I do not assume $f$ is $C^1$.
Is it true that $f$ must be smooth?
This is probably true if we assume $f$ is $C^1$ (see here).
Note: This is false when assuming $f$ is only differentiable almost everywhere, as was shown by Gromov (in fact $M$ can be non-flat, and $N$ flat).
When trying to adapt the argument to this less regular case, we hit a porblem, as described below:
A "proof":
First, note that by the inverse function theorem for everywhere differentiable maps, $f$ is a local homeomorphism.
We would like to prove it's a local isometry w.r.t the intrinsic distances, then use the Myers-Steenrod theorem.
The problem lies within the fine the details: We need to choose suitable length structures on $M,N$ such that $f$ will become an arcwise isometry. We cannot use the class of $C^1$ paths, however, since $f$ does not necessarily map $C^1$ maps to $C^1$ maps.
In a similar spirit, nothing promises us that for a differentiable path $\gamma$ such that $\|\dot \gamma(t)\|$ is integrable, $\|\dot {f \circ \gamma}(t)\|$ will be integrable.
For completeness: The answer is positive. Any differentiable isometry is smooth. This is proved here.