Differentiation of maps Let $f: \mathbb{R}^3 \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ be differentiable. Let $F: \mathbb{R}^2 \to \mathbb{R}$ be defined by the equation
$$F(x,y)=f(x,y,g(x,y)).$$
(a)Find $DF$ in terms of the partials of $f$ and $g$.
(b) If $F(x,y)=0$ for all $(x,y)$, find $D_1g$ and $D_2g$ in terms of the partials of $f$. 
 A: it's easier to see when you define a function $h: R^2 \rightarrow R^3 $ where:
$$h(x,y)=(x,y,g(x,y))=(h_1(x,y),h_2(x,y),h_3(x,y))$$
where $h_i: R^2\rightarrow R, i=1,2,3$, and note that $F(x,y)=f(h(x,y))=(f\,o\,h)(x,y)$, here we can use the chain rule (of course h is differentiable), and...
$$DF(x,y)=D(f\,o\,h)(x,y)=Df(h(x,y))*Dh(h,y)$$
then you only have to compute the last equation, but that's not difficult, then you will be in the result given by Berci
Hope help someone
A: (a) $DF$ is the $1\times 2$ matrix of $(D_1F,D_2F)$ where $D_i$'s denote the partials.
Well,
$$\begin{align} D_1F &=D_1f(x,y,g(x,y)) + D_3f(x,y,g(x,y))\cdot D_1g(x,y)  \\
D_2F &=D_2f(x,y,g(x,y)) + D_3f(x,y,g(x,y))\cdot D_2g(x,y) 
\end{align}$$
I guess the real question is: Why is it addition between the different partials, and.. well.. they are somehow projections of vectors, maybe someone else can give a more explicit explanation on that.
(b) is straightforward, since then $DF=(0,0)$ , so the left hand sides above become $0$, and we get
$$D_ig(x,y) = -\frac{D_if(x,y,g(x,y))}{D_3f(x,y,g(x,y))} . $$
