The two equations $F (x, y, u, v) = 0$ and $G (x, y, u, v) = 0$ determine $x$ and $y$ implicitly as functions of u and v, say $ x = X (u, v) $ and $ y = Y (u, v) $. Show that $$ {\partial X \over \partial u} = {{\dfrac{\partial (F, G)}{\partial (y, u)}} \over \dfrac{\partial (F, G)}{\partial (x, y )}} $$ at the points where the Jacobian $ {\partial (F, G) \over \partial (x, y)} \neq 0 $ and finds similar formulas for the partial derivatives $ {\partial X \over \partial v} , {\partial Y \over \partial u}, {\partial Y \over \partial v} $.
I need help with this exercise, I don't know how to take the function, should I define a composition or another function? Like $h(u,v)=F(x,y,u,v)=F(X(u,v),Y(u,v),u,v)$?
Let $h(u,v):=F(X(u,v),Y(u,v),u,v)$ and $k(u,v):=G(X(u,v),Y(u,v),u,v)$. We know that $h,k$ are identically $0$, so their derivatives are also $0$
Then by the chain rule
$$ \begin{align} 0&=&h_u &=& F_x X_u +F_y Y_u + F_u+0\\ %0&=&h_v &=& F_x X_v +F_y Y_v + 0+ F_v\\ \text{and}\\ 0&=&k_u &=& G_x X_u +G_y Y_u + G_u+0\\ %0&=&k_v &=& G_x X_v +G_y Y_v + 0+ G_v\\ \end{align} $$ which we can write as $$ \begin{pmatrix} F_x & F_y\\ G_x & G_y \end{pmatrix} \begin{pmatrix} X_u\\ Y_u \end{pmatrix} =- \begin{pmatrix} F_u\\ G_u \end{pmatrix}. $$ Inverting the $2\times 2$ matrix we have $$ \begin{pmatrix} X_u\\ Y_u \end{pmatrix} = \frac{1}{\dfrac{\partial (F,G)}{\partial (x,y)}} \begin{pmatrix} -G_y & F_y\\ G_x & -F_x \end{pmatrix} \begin{pmatrix} F_u\\ G_u \end{pmatrix}, $$ and we can then see that $$ X_u = {{\dfrac{\partial (F, G)}{\partial (y, u)}} \over \dfrac{\partial (F, G)}{\partial (x, y )}}, $$ with a similar expression for $Y_u$.
Differentiating $h,k$ with respect to $v$ will give the formulae for $X_v, Y_v$.