I'm studying the Implicit Function Theorem and this is a problem from Munkres' Analysis on Manifolds.
Let $F:\mathbb{R^2} \to \mathbb{R}$ be of class $C^2$, with $F(0,0)=0$ and $DF(0,0)=\begin{bmatrix} 2 & 3\end{bmatrix}$. Let $G:\mathbb{R^3} \to \mathbb{R}$ be defined by the equation $$G(x,y,z)=F(x+2y+3z-1, x^3+y^2-z^2).$$
(a) Note that $G(-2,3,-1)=F(0,0)=0$. Show that one can solve the equation $G(x,y,z)=0$ for $z$, say $z=g(x,y)$, for $(x,y)$ in a neighborhood $B$ of $(-2,3)$, such that $g(-2,3)=-1$.
(b) Find $Dg(-2,3)$.
(c) If $D_1D_1F=3$ and $D_1D_2F=-1$ and $D_2D_2F=5$ at $(0,0)$, find $D_2D_1g(-2,3)$.
I've worked through (a) and (b). This is my answer.
Define $h(x,y,z)=(x+2y+3z-1,x^3+y^2-z^2)$, then $G=F\circ h$, so by Chain Rule, $DG=DF(h(x,y,z))Dh(x,y,z)$.
Since $Dh=\begin{bmatrix} 1 & 2 & 3 \\ 3x^2 & 2y & -2z \\ \end{bmatrix},$ then $DG(-2,3,-1)=\begin{bmatrix} 38 & 22 & 12 \\ \end{bmatrix}$.
So $\frac{\partial G}{\partial z}(-2,3,-1)=12\neq 0$. Hence by the Implicit Function Theorem, we can find such $g$ and $B$ in (a).
Also, we've got $Dg(-2,3)=\frac{-1}{12}\begin{bmatrix} 38 & 22 \\ \end{bmatrix}$ from the result obtained above.
However, I'm lost on how to approach (c). I'm not sure how I'm supposed to find the value. I would appreciate some help on this problem.