Proving $f(\lVert x\rVert)$ is differentiable in $\Bbb R^n$ Let $f: \Bbb R \to \Bbb R$ be an even, differentiable function, 
and let $F: \Bbb R^n \to \Bbb R$ be defined as follows: $F(x) = f(\lVert x\rVert)$
,when $\lVert x\rVert = \sqrt{\sum_{i=1}^n x_i^2}$ .
Prove that $F$ is differentiable in $\Bbb R^n$.
What I tried: It's pretty easy to see that $F$ is differentiable in every $x \neq 0$. For $x = 0$ I got for the partial derivative of $x_1$: 
$\frac{\partial F}{\partial x_1}=\lim_{h\to 0} \frac{F(h,...0)-F(0,..,0)}{h} = 
\lim_{h\to 0} \frac{f(|h|)-f(0)}{h}$ = $f'(0)$
When the last equality follows since $f$ is even. Next I want to show that  $\frac{\partial F}{\partial x_1}$ is continuous in $0$, and then finish, but I couldn't prove it.
**Edit: since $f$ is even and differentiable, we also have $f'(0) = 0$. Still having trouble finishing the proof. 
 A: Using your argument, one can easily show that $\frac{\partial F}{\partial x_i}(0)=f'(0)$ for all $i$, so that
$$\nabla F(0)=(f'(0),f'(0),\dots,f'(0))=f'(0)\cdot (1,1,\dots,1).$$
Now, let $g(x)=\lVert x\rVert$.
For $x\neq 0$, $g$ is differentiable and $g_i(x)=\frac{x_i}{\lVert x\rVert}$, which implies $\nabla g(x)=\frac{x}{\lVert x \rVert}$.
We may calculate $\nabla F(x)$ for $x\neq 0$ using the chain rule.
Let $J$ stand for the Jacobian matrix of a function; for instance, $JF=\nabla F$ and $Jf=f'$ (and here we understand $f'$ as the linear transformation).
Then:
\begin{align}
\nabla F(x)=JF(x)=J\big[  f(g(x)) \big]&=Jf\big(g(x)\big)\cdot Jg(x)
\\&=f'(g(x))\cdot \nabla g(x)
\\&=f'(\lVert x \rVert)\cdot \frac{x}{\lVert x \rVert}
\end{align}
Do we have $\nabla F(x) \to \nabla F(0)$ as $x \to 0$?

The computation above shows that there is something to be asked about the limit of
$$\frac{f'(\lVert x\rVert)}{\lVert x \rVert}\cdot x$$
as $x\to 0$.
I was hoping this would hint towards the insight, but as others have pointed out, we have $f'(0)=0$.
In particular, we have $\nabla F(0)=0$.
Moreover, since $\frac{x}{\lVert x \rVert}$ is bounded, what can you conclude about $\nabla F(x)$ as $x\to 0$?

Let's try to show that $F$ is differentiable at $x=0$, using the definition. We need to show that there is some linear transformation $T:\mathbb{R}^n\longrightarrow\mathbb{R}$ such that
$$\lim_{\lVert h\rVert\to0}\frac{|F(h)-F(0)-Th|}{\lVert h\rVert}=0$$
The obvious candidate for $T$ is $\nabla F (0)=0$, the zero transformation.
Do you think you can take it from here?
A: Since $f$ is even we know $f(-x) = f(x)$. This will be very useful when we regard the first derivative of $f$ at $0$. Since $f$ is differentiable everywhere $f'(0) =  \lim_{x\to 0} \frac{f(x)-f(0)}{x}$ exists. So both the right and the left sided limes exists which leads to
\begin{align}
\lim_{x\to 0+} \frac{f(x)-f(0)}{x} &= f'(0)
=\lim_{x\to 0-} \frac{f(x)-f(0)}{x} 
= \lim_{t\to 0+} \frac{f(-t)-f(0)}{-t} \\
&= -\lim_{t\to 0+} \frac{f(t)-f(0)}{t}
.
\end{align}
Therefore $\lim_{x\to 0+} \frac{f(x)-f(0)}{x}$ must be $0$ and consequently $f'(0)=0$.
This fact will be very useful to show differentiability of $F$ at $0$.
