= Solution
<Differentiability> at $x_0$ means that a <linear map> $A:\mathbb R^n\to\mathbb R^m$ exists with
$$
f(x_0+h)=f(x_0)+Ah+o(\|h\|).
$$
The map $A=Df(x_0)$ is the <total derivative>. A <linear map> between these finite-dimensional spaces is bounded, so $\|f(x_0+h)-f(x_0)\|\leq\|A\|\|h\|+o(\|h\|)\to0$. Thus <differentiability> implies <continuity>.
For the displayed two-variable function,
$$
0\leq\frac{x^4}{\sqrt{x^4+y^2}}\leq x^2\leq x^2+y^2.
$$
Writing $r=\sqrt{x^2+y^2}$ gives $|f(x,y)|/r\leq r\to0$. Therefore \b[it is differentiable at the origin, with derivative zero].
Finally, put $h=x-x_0$, write $f(x_0+h)=Ah+r(h)$ with $r(h)=o(\|h\|)$, and $g(x_0+h)=g(x_0)+s(h)$ with $s(h)=o(1)$. Since $f(x_0)=0$,
$$
F(x_0+h)=\langle Ah,g(x_0)\rangle
+\langle Ah,s(h)\rangle+\langle r(h),g(x_0)+s(h)\rangle.
$$
By the <Cauchy-Schwarz inequality> the last two terms are $o(\|h\|)$: $A$ is bounded and $g(x_0)+s(h)$ stays bounded. Hence
$$
\boxed{DF(x_0)h=\langle Df(x_0)h,g(x_0)\rangle.}
$$
Only <continuity> of $g$ at $x_0$ is needed, because the vanishing of $f(x_0)$ removes the term which would otherwise require differentiating $g$.
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