This is a self-answered question. I post it here since it wasn't obvious to me. It is a follow-up of this previous question of mine.
Let $U \subseteq \mathbb{R}^n$ be an open set containing the origin, and let $h:[0,\infty] \to [0,\infty)$ be a smooth function satisfying $h(0)=0$.
Define $f:U \to \mathbb R^n$ by $ f(x)=
\begin{cases} \frac{h(\|x\|)}{\|x\|}x & \text{if $x\neq 0$} \\ 0 & \text{if $x=0$}\end{cases}$.
If we describe a point $x \in \mathbb R^n$ using its distance from the origin $r=\|x\| \in \mathbb R_{\ge 0}$, and its "direction" $\sigma = \frac{x}{\|x\|} \in \mathbb S^{n-1}$, then $f$ is given by
$(r,\sigma)\to (h(r),\sigma)$.
Proposition: $f$ is smooth if and only if $h$ is smooth and $h^{2k}(0)=0$ for every natural $k$.
In particular, $f$ is smooth whenever $h$ is an odd smooth function. Moreover, if $h$ is real-analytic, then $f$ is smooth if and only if $h$ is odd. (Since for real-analytic functions the condition on the derivatives implies being odd).
It is not hard to see that the vanishing of all the derivatives of even order is necessary. Indeed, the map $$x \mapsto f(x,0,0,\dots,0)=\frac{h(|x|)}{|x|}(x,0,0,\dots,0)=\big((h(|x|)\text{sgn}(x),0,0,\dots,0\big)$$
should be smooth. Now, it is not hard to see that the function $x \to h(|x|)\text{sgn}(x)$ is infinitely differentiable at zero if and only if $h^{2k}(0)=0$ for every natural $k$.
Best Answer
Proof:
The idea is to reduce the question to a "scalar-valued" case, i.e. to understand when smoothness of a function $x \mapsto \phi(x)$ implies the smoothness of $x \mapsto \phi(\|x\|)$.
First, note that $f$ can be rewritten as $f(x)=F(x) \cdot x$, where $F:U \to \mathbb R$ is given by
$$F(x) = \begin{cases} \frac{h(\|x\|)}{\|x\|} & \text{if $x\neq 0$} \\ h'(0) & \text{if $x=0$}\end{cases}. $$
Thus, it suffices to prove that $F$ is smooth.
Define $g:[0,\infty) \to \mathbb R$ by
$$g(x) = \begin{cases} \frac{h(x)}{x} & \text{if $x\neq 0$} \\ h'(0) & \text{if $x=0$}\end{cases} $$
Now, $h \in C^{\infty}$ together with $h(0)=0$, imply that $g \in C^{\infty}$, and $g^{m}(0)=\frac{h^{(m+1)}(0)}{m+1}$ for every natural $m$. Thus, all the odd derivatives of $g$ vanish at zero.
Now, this answer implies that the map $x \mapsto g(\|x\|)=F(x)$ is smooth, as required.