Is there a function $f\colon \Bbb R\to \Bbb R$ in $C^\infty$ such that the set $$D=\{y\in \operatorname{Im}(f):\exists x\in f^{-1}(y)\text{ with }f'(x)=0\}$$ is uncountable?
This question arose when thinking about the regular values of a smooth map into a differentiable manifold. Sard's Theorem says that $D$ has measure zero, so a candidate here will probably be exotic.
It is not too hard to construct such a function where $D$ is dense: take a listing of the rationals and a function which has a plateau between $n$ and $n+1$ at the height of the $n$-th rational.
I will preemptively warn that this question is not about finding a nontrivial function whose derivative vanishes on an uncountable set. Such a function can be constructed by requiring $f(x)=0$ on a cantor set and extending smoothly.
Let $f(x)$ be a $C^\infty$ function which vanishes on the Cantor set and is positive everywhere else, e.g., the function constructed by Nate Eldredge in this Math Overflow answer. Then the function $$g(x)=\int_0^xf(t)dt$$ is a strictly increasing $C^\infty$ function whose derivative vanishes precisely on the Cantor set, whence its set $D$ of critical values is a bijective image of the Cantor set and so has cardinality $\mathfrak c.$