The goal of this question is to isolate some aspects of this geometric question.
Let $\psi:[0,\infty) \to [0,\infty)$ be a smooth strictly increasing function satisfying $\psi(0)=0$ and let $\phi:(0,\infty) \to \mathbb R$ be smooth.
Suppose that $B:=\lim_{x \to 0^+}\phi'(x)\psi(x)$ is finite.
Define $$f_1(x)=\psi(x)\cos(\phi(x)), \, \,\,\,f_2(x)=\psi(x)\sin(\phi(x)).$$ $f_i$ are clearly defined on the open interval $(0,\infty)$; we extend them continuously to zero by setting $f_i(x)=0$. (Recall that $\psi(0)=0$).
Can both the $f_i$ be infinitely (right) differentiable at $x=0$?
So far I managed to derive a necessary condition for the $f_i$ to be $C^1$ at zero, under the additional assumption that $\psi'(0)>0$. I am not sure how to proceed to check higher derivatives in a tractable way, or what happens when $\psi'(0)=0$.
The necessary condition: I prove below that we must have $B=0$ and that $\lim_{x \to 0^+} \phi(x)$ exists.
Indeed,
$$ \frac{f_1(x)-f_1(0)}{x}=\frac{\psi(x)-\psi(0)}{x}\cos(\phi(x))\Rightarrow \\ \cos(\phi(x))=\frac{f_1(x)-f_1(0)}{x} \frac{1}{\frac{\psi(x)-\psi(0)}{x}} \Rightarrow \\ \lim_{x \to 0^+} \cos(\phi(x))=\frac{f_1'(0)}{\psi'(0)}, $$
and similarly $\lim_{x \to 0^+} \sin(\phi(x))=\frac{f_2'(0)}{\psi'(0)}$.
Thus, both $\lim_{x \to 0^+} \cos(\phi(x)), \lim_{x \to 0^+} \sin(\phi(x))$ exist, and hence so does $\lim_{x \to 0^+} \phi(x)$.
Now, a direct calculation shows that $$ f_1'(x)=\begin{cases} \psi'(x)\cos(\phi(x))-\psi(x)\phi'(x)\sin(\phi(x)) & \text{if $x > 0$} \\ \psi'(0)\cdot \lim_{x \to 0^+} \cos(\phi(x)) & \text{if $x=0$}\end{cases}$$
In particular, $$ \lim_{x \to 0^+} f_1'(x)=\psi'(0)\cdot \lim_{x \to 0^+}\cos(\phi(x))-B\lim_{x \to 0^+} \sin(\phi(x))$$
Since we assumed $f_1$ is $C^1$ at zero, we must have $B\lim_{x \to 0^+} \sin(\phi(x))=0$.
Applying the same considerations for $f_2$, we deduce that $B\lim_{x \to 0^+} \cos(\phi(x))=0$. Since $\sin$ and $\cos$ cannot converge both to zero at the same point, we conclude that $B=0$.
Here's an example, without the assumption $\psi'(0)>0$:
$$\psi(x)=e^{-\frac{1}{x^2}}\\ \phi(x)=\frac{1}{x}$$