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 $v$ 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
$h(u,v)=F(x,y,u,v)=F(X(u,v),Y(u,v),u,v)$?
 A: 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 $X_v, Y_v$.
