calculus - Chain rule for two variables and finding $Df(x)(a)$ of $f(x)=arctg(||x||^2)*x$

I worked on one problem I stuck because I think I have a problem about applying chain rule at function with more than one variable. I will write what I try to do, but please explain my step by step so that can I understand what differental operators in chain rule do.
$$f(x)=arctg(||x||^2)*x$$
Find a $Df(x)(a)$ if $x = (1, 1, ..., 1)$ and $a = (1, 2, ..., n)$.
My work:
It is easy to see that we will need chain rule and product rule to solve this problem. First of all let's look at $arctg(||x||^2)$.

This function we can write as $arctg\circ n$. Where $n$ is $n(x)=||x||^2$. I write in that way because we know $||x||^2=$and we can easily compute derivate of n(product rule).$Dn(x)(h)=2$. And for$arctg(x)$we know that derivative of$arctg(x)$is$\frac{1}{1+x^2}$.

But I have problem in using chain rule...

Answer

$f = \phi \circ (\arctan \circ \|\cdot\|^2) = \phi \circ g$, where $\phi: t \in \Bbb R \mapsto \phi(t) = tx \in \Bbb R^n$.

We have $Dg(x) = D(\arctan)(\|x\|^2) D(\|\cdot\|^2)(x)$, i.e. for each $h$,

$$Dg(x)h = \frac{2\langle x,h\rangle}{1 + \|x\|^4}$$

Also, as $\phi$ is linear and continuous, $D\phi(t) = \phi$. Thus: $Df(x) = D\phi(g(x)) \circ Dg(x) = \phi \circ Dg(x)$. In other words, for each $h$:

$$Df(x)h = \phi(Dg(x)h) = \frac{2\langle x,h\rangle}{1+\|x\|^4} x$$