Analysis on Manifolds problem 9.4 (Implicit function theorem) From Analysis on Manifolds by  Munkres.
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 $D_2D_1g(-2,3)$.
 A: 
There is a comment, but I still do not understand what the D1D2F symbol means in this context.

I never met such notation before but I guess that since $F$ is a function of two variables (say, $x_1$ $x_2$) then $D_1D_2$ denotes the partial derivative $\frac{\partial^2 F}{\partial x_2\partial x_1}$. Under some assumptions it equals to $\frac{\partial^2 F}{\partial x_1\partial x_2}$. Namely, the following theorem holds (see, for instance, [Fich, 190])
Theorem. Let $D$ be an open domain of $\Bbb R^2$ and $F:D\to\Bbb R$ be a function such that all partial derivatives $\frac{\partial F}{\partial x_1}$, $\frac{\partial F}{\partial x_2}$, $\frac{\partial^2 F}{\partial x_1\partial x_2}$, and $\frac{\partial^2 F}{\partial x_2\partial x_1}$ exist in $D$ and two latter derivatives are continuous at some point $(y_1,y_2)$ of $D$. Then
$$\frac{\partial^2 F}{\partial x_1\partial x_2}(y_1,y_2)=\frac{\partial^2 F}{\partial x_2\partial x_1}(y_1,y_2).$$
References
[Fich] Grigorii Fichtenholz, Differential and Integral Calculus, vol. I, 5-th edition, M.: Nauka, 1962 (in Russian).
