= Solution
For <differentiable maps> $F$ at $x$ and $G$ at $F(x)$, the <chain rule> asserts
$$
D(G\circ F)_x=DG_{F(x)}\circ DF_x.
$$
Write $F(x+h)=F(x)+Ah+r(h)$ with $\|r(h)\|=o(\|h\|)$, and $G(F(x)+u)=G(F(x))+Bu+s(u)$ with $\|s(u)\|=o(\|u\|)$. Since $u=Ah+r(h)=O(\|h\|)$, substitution gives
$$
G(F(x+h))=G(F(x))+BAh+Br(h)+s(Ah+r(h)).
$$
Both remaining terms are $o(\|h\|)$, proving <differentiability> and the stated <chain rule>. The estimate for $s$ also holds when its argument is zero, by setting $s(0)=0$.
For the circle constraint take $H(u)=u_1^2+u_2^2$. The composite $H\circ F$ is constant, so the <chain rule> gives
$$
0=D(H\circ F)_x=2F(x)^TDF_x.
$$
Since $\|F(x)\|=1$, this is a nonzero row vector annihilating $DF_x$. Thus $DF_x$ cannot be invertible: \b[$\det DF_x=0$ at every $x$]. Geometrically, the <derivative of a map into a level set> takes values in the <tangent space> to the circle.
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