Suppose that $F\left ( x,y,u,v \right )$ and $G\left ( x,y,u,v \right )$ have continuous first partial derivatives. Suppose that the equations $F\left ( x,y,u,v \right )=0$ and $G\left ( x,y,u,v \right )=0$ can be solved for $x, y$ as function of $u, v$.
Express $\frac{\partial \left ( x,y \right ) }{\partial \left ( u,v \right ) } $ in terms of $\frac{\partial \left ( F,G \right ) }{\partial \left ( u,v \right ) } $ and $\frac{\partial \left ( F,G \right ) }{\partial \left ( x,y \right ) } $, where $$\tfrac{\partial \left ( x,y \right ) }{\partial \left ( u,v \right ) }= \begin{vmatrix} x_{u} &x_{v} \\ y_{u} &y_{v} \end{vmatrix}, \qquad \tfrac{\partial \left ( F,G \right ) }{\partial \left ( u,v \right ) }= \begin{vmatrix} F_{u} &F_{v} \\ G_{u}&G_{v} \end{vmatrix}, \qquad\tfrac{\partial \left ( F,G \right ) }{\partial \left ( x,y \right ) }= \begin{vmatrix} F_{x} &F_{y} \\ G_{x}&G_{y} \end{vmatrix} $$ are Jacobian determiants.
Can I solve this problem in the following way?
Consider the function $$H: \mathbb R^4\to \mathbb R^2, \qquad \left ( x,y,u,v \right ) \mapsto (F\left ( x,y,u,v \right ),G\left ( x,y,u,v \right ) )$$
Since $x, y$ can be solved as function of $u, v$ - let's say $$x=x(u,v),\space y=y(u,v),$$ I guess the implicit function theorem can be used here.
Then we have $$DH=\begin{bmatrix} F_x& F_y& F_u& F_v\\ G_x& G_y& G_u&G_v \end{bmatrix},$$in which $$\begin{vmatrix} F_x& F_y \\ G_x& G_y \end{vmatrix}=\tfrac{\partial \left ( F,G \right ) }{\partial \left ( x,y \right ) } \ne 0.$$
So, by the implicit function theorem $$\begin{bmatrix} x_u& x_v\\ y_u& y_v\end{bmatrix}=-\begin{bmatrix} F_x& F_y\\ G_x&G_y\end{bmatrix}^{-1} \times \begin{bmatrix} F_u& F_v\\ G_u&G_v\end{bmatrix}.$$
Finally, we have $$\tfrac{\partial \left ( x,y \right ) }{\partial \left ( u,v \right ) }=-\left[\tfrac{\partial \left ( F,G \right ) }{\partial \left ( x,y \right ) }\right]^{-1}\space \times \space \tfrac{\partial \left ( F,G \right ) }{\partial \left ( u,v \right ) }.$$
Did I do this correctly?