= Solution
<Differentiability> at $x_0$ means that a <linear map> $L:\mathbb R^n\to\mathbb R^m$ exists such that $f(x_0+h)=f(x_0)+Lh+o(\|h\|)$. Since finite-dimensional <linear maps> are <bounded>, the right side tends to $f(x_0)$ as $h\to0$, 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 $r(h)=o(\|h\|)$, and $g(x_0+h)=g(x_0)+s(h)$ with $s(h)=o(1)$. Since $f(x_0)=0$,
$$
F(x_0+h)-F(x_0)-\langle Lh,g(x_0)\rangle=\langle Lh,s(h)\rangle+\langle r(h),g(x_0)+s(h)\rangle=o(\|h\|).
$$
<Cauchy sequence>[Cauchy]–Schwarz and <boundedness> of $g$ near $x_0$ justify the last estimate. Thus $\boxed{dF(x_0)h=\langle df(x_0)h,g(x_0)\rangle}$; <differentiability> of $g$ is unnecessary because $f(x_0)=0$.
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