Let $f\colon \mathbb R \to \mathbb R$ be a function with a certain regularity:
$$f\in C^k(\mathbb R)\quad\text{ for $k \in \mathbb N\cup\{\infty\}$}.$$
Let's consider $\mathscr G$ the graph of $f$:
$$\mathscr G:=\{(x,y)\in \mathbb R^2,\ y=f(x)\}.$$
To test if a graph in $\mathbb R^2$ is a function of $x$, one can use the vertical line test.
Let's assume that we rotate the graph $\mathscr G$ around the origin of a certain angle $\alpha$, such that $r_\alpha(\mathscr G)$ passes the vertical line test, i.e. it is still the graph of a function; let's call it $f_\alpha$.
I am wondering about the following questions:
Can $f_\alpha$ not be in $C^k(\mathbb R)$ anymore?
Can $f_\alpha$ be in $C^{k+1}$?
I think both questions can be answered positively, since you can consider the function $$x\mapsto \begin{cases} \frac{\vert x\vert}x \sqrt{\vert x\vert} &\text{ if $x\ne 0$} \\ 0&\text{ otherwise}\end{cases}$$
which will not be $C^1$ since it is not differentiable at $x=0$, but it can get $C^1$ if you rotate it.
So my question is the following:
Is it true that the regularity will not change outside a discret set of points?
Yes, the complication you demonstrated is the only one.
It follows from the implicit function theorem since $\psi(x,y) = f(x) - y$ has the same regularity as $f$ and this is invariant under rotation and we see as long as the gradient is not parallell to the $x$-axis we have a function solution to $\psi=0$ that is as regular as $\psi$ in an neighborhood.
The complication when the tangent is vertical can not happen in an open set, but it may happen that the points where this happen can have accumulation points.