What conditions on $f$ make $f\circ |\cdot |$ a smooth function? Let $|\cdot |$ be the Euclidean norm on $\mathbb{R}^n$. What conditions (preferably necessary) on $f:[0,\infty)\mapsto\mathbb{R}$ ensure that $f\circ |\cdot |\in C^{\infty}(\mathbb{R}^n)$? For $n=1$, I believe that $f\in C^{\infty}([0,\infty))$ along with $f^{(k)}(0)=0$ for odd $k$ suffice.
 A: It is enough to consider the case $n=1$ by putting all coordinates, except one, equal to zero.
We can assume, without loss of generality, that $f(0)=0$. Then we have that $f(|x|)$ is even and $C^{\infty}$. It is a result of Whitney (Differentiable even functions, Duke Math. J. 10 (1943), 159-160.) that then there is $g\in C^{\infty}$ such that $f(|x|)=g(x^2)$. Since for $x\geq0$ we have $f(|x|)=f(x)=g(x^2)$, this is the definition of $f$.
In particular we can deduce that your condition is necessary. We can just compute derivatives of $f$ from the right. and we get $D_{+}^{n}f(x)=D_{+}^{n}f(|x|)=D^ng(x^2)$.
We can compute that
\begin{align}
Dg(x^2)&=2xg'(x^2)\\
D^2g(x^2)&=2g'(x^2)+4x^2g''(x^2)\\
D^3g(x^2)&=10xg''(x^2)+8x^3g'''(x^2)\\
\ldots&
\end{align}
We see that for $n$ odd $D^ng(x^2)|_{x=0}=0$. If the $\ldots$ are not convincing the Faa di Bruno formula allows to compute the $n$-th derivatives of $g(x^2)$ and see it is multiple of $x$ for $n$ odd. Therefore $D_{+}^{2k+1}f(0)=0$ is a necessary condition.
So, the necessary and sufficient condition is that $f(x)=g(x^2)$ for some $C^{\infty}$ function $g$. In other words, to get rid of the square root, you have to square.
