 Solution

Differentiability at $x_0$ means that a linear map $A:{\mathbb{R}}^n\to {\mathbb{R}}^m$ exists withThe map $A=Df(x_0)$ is the total derivative. A linear map between these finite-dimensional spaces is bounded, so $\Vert f(x_0+h)-f(x_0)\Vert \le \Vert A\Vert \Vert h\Vert +o(\Vert h\Vert )\to 0$. Thus differentiability implies continuity.

For the displayed two-variable function,Writing $r=\sqrt{x^2+y^2}$ gives $|f(x,y)|/r\le r\to 0$. Therefore it is differentiable at the origin, with derivative zero.

Finally, put $h=x-x_0$, write $f(x_0+h)=Ah+r(h)$ with $r(h)=o(\Vert h\Vert )$, and $g(x_0+h)=g(x_0)+s(h)$ with $s(h)=o(1)$. Since $f(x_0)=0$,By the Cauchy-Schwarz inequality the last two terms are $o(\Vert h\Vert )$: $A$ is bounded and $g(x_0)+s(h)$ stays bounded. HenceOnly continuity of $g$ at $x_0$ is needed, because the vanishing of $f(x_0)$ removes the term which would otherwise require differentiating $g$.