 Solution

Differentiability at $x_0$ means that a linear map $L:{\mathbb{R}}^n\to {\mathbb{R}}^m$ exists such that $f(x_0+h)=f(x_0)+Lh+o(\Vert h\Vert )$. Since finite-dimensional linear maps are bounded, the right side tends to $f(x_0)$ as $h\to 0$, proving continuity.

For the specified scalar function, $0\le f(x,y)\le x^2\le x^2+y^2$. Dividing by $\sqrt{x^2+y^2}$ gives a bound tending to zero at the origin. Consequently $\boxed{df(0,0)=0}$ and the function is differentiable there.

For the final product assertion, write $f(x_0+h)=Lh+r(h)$, where $r(h)=o(\Vert h\Vert )$, and $g(x_0+h)=g(x_0)+s(h)$ with $s(h)=o(1)$. Since $f(x_0)=0$,Cauchy–Schwarz and boundedness of $g$ near $x_0$ justify the last estimate. Thus $\boxed{dF(x_0)h=⟨df(x_0)h,g(x_0)⟩}$; differentiability of $g$ is unnecessary because $f(x_0)=0$.