Show that $F(x) = f(\|x\|)$ is differentiable on $\mathbb{R}^n$. 
Let an even function $f:\mathbb{R}\to\mathbb{R}$ which is even and differentiable. We define $F:\mathbb{R}^n\to\mathbb{R}$  as $F(x) = f(\|x\|)$. Show that $F(x)$ is differentiable on $\mathbb{R}^n$.

My Work: 
Let $x_0\in\mathbb{R}^n$.


*

*If $x_0\ne 0$ then the partial derivatives of the norm are well-defined and:
$$ \frac{\partial \|.\|}{\partial x_i} = \frac{x_i}{\|x\|}$$
$\frac{x_i}{\|x\|}$ continuous and since $f$ is differentiable it's partial derivatives are continuous. So we can conclude that $\frac{\partial F}{\partial x_i}$ is continuous and therefore $F$ is differentiable.

*If $x_0=0$ then the $\|x_0\| = 0$ and since $f$ is even $f(0)=0$ and it must be an extremum point. Therefore, $F'(0)=0$.


I'd like to get a critique of my work. Am I being right/rigorous?
Thanks. 

While the case $x_0 \neq 0$ is correct and trivial, your proposed solution for the case $x_0=0$ is not a proof at all. A function can have a maximum or minimum at a point where it is not smooth (e.g. $x \mapsto |x|$ in $\mathbb{R}$).
Here you should use the fact that $f'(0)=0$ (because $f$ is even and differentiable at zero), so that
$$
f(t)=f(0)+o(t) \quad\hbox{as $t \to 0$}. \tag{1}
$$
Now plug $t=\|x\|$ into $(1)$ and deduce that
$$
F(x)=F(0)+o(\|x\|) \quad\hbox{as $x \to 0$},
$$
which shows that $DF(0)=0$.
