Differential calculation in multiple variables function (cannot understand 2nd order differential form) This question is somehow related to this question. 
Consider a multiple variables function $G(u, v) = \left(\matrix{x(u,v) &=& G_x(u,v)\\y(u,v) &=& G_y(u,v)\\z(u,v) &=& G_z(u,v)}\right)$. 
I am trying to write " $\frac{\partial z}{\partial x}$ " and " $\frac{\partial z}{\partial y}$ ". 
Considering that : 
$$\mathrm{d}G_x = \frac{\partial G_x}{\partial u}\mathrm{d}u + \frac{\partial G_x}{\partial v}\mathrm{d}v$$
$$\mathrm{d}G_y = \frac{\partial G_y}{\partial u}\mathrm{d}u + \frac{\partial G_y}{\partial v}\mathrm{d}v$$
$$\mathrm{d}G_z = \frac{\partial G_z}{\partial u}\mathrm{d}u + \frac{\partial G_z}{\partial v}\mathrm{d}v$$
I make the bet that I am looking for $$\frac{\mathrm{d}G_z}{\mathrm{d}G_x}\bigg|_{\mathrm{d}G_y=0}$$
using 
$\mathrm{d}G_y = 0$, 
I get 
$$\frac{\mathrm{d}G_y}{\partial u}\mathrm{d}u = -\frac{\mathrm{d}G_y}{\partial v}\mathrm{d}v$$
so that, eliminating ${\rm d}u$ and ${\rm d}v$ using the previous equation : 
$${\frac{\partial z}{\partial x} = \frac{\mathrm{d}G_z}{\mathrm{d}G_x} = 
\frac{\frac{\partial G_z}{\partial u} \frac{\partial G_y}{\partial v} - \frac{\partial G_y}{\partial u} \frac{\partial G_z}{\partial v}}{\frac{\partial G_x}{\partial u} \frac{\partial G_y}{\partial v} - \frac{\partial G_y}{\partial u} \frac{\partial G_x}{\partial v}}}$$
I don't know much about multiple variables differentiation, so can somebody tell me if this reasoning is OK?
If the reasoning is OK, how to build a similar reasoning to write $\frac{\partial^2 z}{{\partial x}^2}$ ?
What I cannot get is how to write a similar constraint as $\mathrm{d}G_y = 0$ for the 2nd order differential form so that I can get equations of ${\rm d}u^2$, ${\rm d}v^2$ and ${\rm d}u{\rm d}v$. 
 A: I have found a way to solve my problem : 
Instead of writing differential forms, it is also possible to simply differentiate the function $z = f(x, y) = f\left(x(u, v), y(u,v)\right)$ assuming that such a function exists. So I am looking for 
$$
\frac{\partial z}{\partial x} = \frac{\partial f}{\partial x} 
\quad{\rm and }\quad
\frac{\partial z}{\partial y} = \frac{\partial f}{\partial y} 
$$
First differentiate along $u$ : 
$$
\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial u}
$$
Do the same along $v$ :
$$
\frac{\partial z}{\partial v} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial v}
$$
So we obtain a $2\times2$ linear system of $\frac{\partial z}{\partial x} 
$ and $\frac{\partial z}{\partial y}$. The solution of this system is the same as the one in the question...
Using this method and derivative of composition functions formula, one one can write a $3\times 3$ linear system of $\frac{\partial^2 z}{\partial x^2}$, $\frac{\partial^2 z}{\partial y^2}$ and $\frac{\partial^2 z}{\partial x\partial y}$. Solving this system I get my solution.
I have checked numerically these solutions, it seems to be OK.
