= Solution
<Differentiability> at $p$ means there is a <linear map> $A=Df|_p$ such that
$$
f(p+h)=f(p)+Ah+r(h),\qquad
\frac{\|r(h)\|}{\|h\|}\longrightarrow0\quad(h\to0).
$$
This is the <Fréchet derivative> definition; the same linear approximation must work in every direction, not merely along coordinate axes.
The <chain rule> says that if $f$ is <differentiable> at $p$ and $g$ is <differentiable> at $f(p)$, then $g\circ f$ is <differentiable> at $p$ and
$$
\boxed{D(g\circ f)|_p=Dg|_{f(p)}\circ Df|_p.}
$$
For its proof, put $A=Df|_p$, $B=Dg|_{f(p)}$, and $k=f(p+h)-f(p)=Ah+r(h)$. Since a <linear map> between these finite-dimensional normed spaces is bounded, $\|k\|=O(\|h\|)$. <Differentiability> of $g$ gives $g(f(p)+k)=g(f(p))+Bk+s(k)$ with $s(k)=o(\|k\|)$. Therefore
$$
g(f(p+h))-g(f(p))-BAh=Br(h)+s(k)=o(\|h\|).
$$
If $k=0$, take $s(0)=0$; otherwise $\|s(k)\|/\|h\|=(\|s(k)\|/\|k\|)(\|k\|/\|h\|)\to0$. This proves the required full linear approximation.
For the <Euclidean norm>, at every $x\ne0$ the ordinary one-coordinate differentiation gives
$$
\boxed{\frac{\partial\|x\|}{\partial x_i}=\frac{x_i}{\|x\|}.}
$$
At the origin its difference quotient in direction $e_i$ is $|t|/t$, whose right and left limits are $1$ and $-1$. Thus \b[none of its coordinate <partial derivatives> exists at the origin].
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