Given a function $z:\Bbb R^2 \rightarrow\Bbb R$ and $F:\Bbb R^3 \rightarrow \Bbb R$ (both are $C^2$) and knowing that $F(x,y,z(x,y))=0$ and $\frac{ \partial F}{\partial z}(x,y,z(x,y))\ne 0$ find $\frac{ \partial^2 z}{\partial x \partial y}$.
I think I did this but I would really appreciate it if somebody checked if I'm right as I struggled with it for quite some time.
$\frac{ \partial F}{\partial x}\frac{\partial x}{\partial x}+\frac{ \partial F}{\partial y}\frac{\partial y}{\partial x}+\frac{ \partial F}{\partial z}\frac{\partial z}{\partial x}=\frac{ \partial F}{\partial x}+\frac{ \partial F}{\partial z}\frac{\partial z}{\partial x}=F_x'+F_z'z_x'=0$
$\frac{\partial }{\partial y}(F_x'+F_z'z_x')=\frac{\partial F_x'}{\partial x}\frac{\partial x}{\partial y}+\frac{\partial F_x'}{\partial y}\frac{\partial y}{\partial y}+\frac{\partial F_x'}{\partial z}\frac{\partial z}{\partial y}+z_x'(\frac{\partial F_z'}{\partial x}\frac{\partial x}{\partial y}+\frac{\partial F_z'}{\partial y}\frac{\partial y}{\partial y}+\frac{\partial F_z'}{\partial z}\frac{\partial z}{\partial y})+F_z'z_{xy}''=F_{xy}''+F_{xz}''z_y'+z_x'(F_{zy}''+F_{zz}''z_y')+F_z'z_{xy}''=0$
so: $z_{xy}''=-\frac{F_{xy}''+F_{xz}''z_y'+z_x'(F_{zy}''+F_{zz}''z_y')}{F_z'}$, where $z_x'=-\frac{F_x'}{F_z'}$ and $z_y'=-\frac{F_y'}{F_z'}$