The question says:
Assume that $f: \mathbb R^3 \to \mathbb R$ and $g: \mathbb R^2 \to \mathbb R$ are differentiable and $h: \mathbb R^2 \to \mathbb R$, $h(x,y)=f(x,y,g(x,y))$.
(a) Write D$h(x,y)$ in terms of the partial derivatives of $f$ and $g$.
(b) Suppose that $h(x,y)=0$ $\forall (x,y) \in \mathbb R^2$. Write $\frac {\partial g} {\partial x}(x,y)$ and $\frac {\partial g} {\partial y}(x,y)$ in terms of the partial derivatives of $f$ at the point $(x,y,g(x,y))$.
EDIT: I think I have figured out part (a)... but now I'm stuck with part (b). Can't find a way to write $\frac {\partial g} {\partial x}(x,y)$ and $\frac {\partial g} {\partial y}(x,y)$ or find out what difference it makes that $h(x,y)=0$. If anybody has any hints on how to approach part (b), please help me! Thanks
If $F : \mathbb R^n \to \mathbb R^m$, let's say $F(x_1,\dots,x_n) = \big( F_1(x_1,\dots,x_n), \dots, F_m(x_1,\dots,x_n) \big)$, is a differentiable function, the derivative of $F$ at a point $a \in \mathbb R^n$ is the $m \times n$ matrix ${\rm D}F(a)$ whose $i$-th row contains all the partial derivatives of $F_i$ at $a$, that is, the $i$-th row of ${\rm D}F(a)$ is $$\big[ \partial_1F_i(a) \ \ \cdots \ \ \partial_nF_i(a) \big].$$ In your case, if you write $h$ as $f \circ j$, where $j : \mathbb R^2 \to \mathbb R^3$ is the function $j(x,y) = (x,y,g(x,y))$, by the chain rule you have $$\begin{align} {\rm D}h(x,y) &= {\rm D}f(j(x,y)) {\rm D}j(x,y) \\ &= \begin{bmatrix} \partial_1f(j(x,y)) & \partial_2f(j(x,y)) & \partial_3f(j(x,y)) \end{bmatrix} \begin{bmatrix} 1&0 \\ 0&1 \\ \partial_1g(x,y) & \partial_2g(x,y) \end{bmatrix} \\ &= \begin{bmatrix} \partial_1f(j(x,y)) + \partial_3f(j(x,y)) \partial_1g(x,y) & \partial_2f(j(x,y)) + \partial_3f(j(x,y)) \partial_2g(x,y) \end{bmatrix}. \end{align}$$ So, if $h=0$, $$\begin{align} {\rm D}h(x,y)=[0 \quad 0] &\implies \begin{cases} \partial_1f(j(x,y)) + \partial_3f(j(x,y)) \partial_1g(x,y) = 0 \\ \partial_2f(j(x,y)) + \partial_3f(j(x,y)) \partial_2g(x,y) = 0 \end{cases} \\[1mm] &\implies \begin{cases} \partial_1g(x,y) = - \dfrac{\partial_1f(j(x,y))}{\partial_3f(j(x,y))} \\ \partial_2g(x,y) = - \dfrac{\partial_1f(j(x,y))}{\partial_3f(j(x,y))} \end{cases} \end{align}$$