Q:

Given $x+y=uv$ and $xy=u-v$ s.t. $x = X(u,y), v = V(u,y),$ find the partials of $X$ and $V$ w.r.t $u$ and $y$.

The equations $x+y=uv$ and $xy=u-v$ determine $x$ and $v$ implicitly as functions of $u$ and $y$ such that $x = X(u,y), v = V(u,y).$ Find the partial derivatives of $X$ and $V$ with respect to $u$ and $y$.
I tried taking the partial derivatives of the two equations with respect to $u$ and got $$\frac{\partial X}{\partial u} + \frac{\partial y}{\partial u} = v,$$ $$\frac{\partial X}{\partial u}y + \frac{\partial y}{\partial u}x = 1.$$ However, these equations don't yield the correct value of $\frac{\partial X}{\partial u}$. 
Also, I thought it was valid to fix $y$ in the above scenario so that $\frac{\partial y}{\partial u}=0$, but that implies that $\frac{\partial X}{\partial u} = v$, which is also false. Can someone please explain what I'm doing wrong? 

A:

The equations are
\begin{equation}
x+y = uv, \quad xy = u-v.
\end{equation}
They can be solved for $v$ and $x$ by multiplying the first equation with $y$: $xy + y^2 = u v y$, and by using the second equation to eliminate $x$:
\begin{equation}
u-v + y^2 = xy + y^2 = uvy \quad \Leftrightarrow \quad v = \frac{u+y^2}{1+uy} =: V(u,y).
\end{equation}
We also obtain
\begin{equation}
x = uv - y = u \frac{u+y^2}{1+uy} - y = \frac{u^2-y}{1+uy} =: X(u,y).
\end{equation}
We can now compute the partial derivatives of $V, X$ by the quotient rule:
\begin{equation}
\frac{\partial V}{\partial u} = \frac{1-y^3}{(1+uy)^2}, \quad \frac{\partial V}{\partial y} = \frac{2y+uy^2-u^2}{(1+uy)^2}, \quad \frac{\partial X}{\partial u} = \frac{2u+u^2y+y^2}{(1+uy)^2}, \quad \frac{\partial X}{\partial y} = \frac{-1-u^3}{(1+uy)^2}.
\end{equation}
Attempting implicit differentiation, on the other hand, we do not start by solving the equations. Instead we take the partial derivatives with respect to $u$ and $y$ of the equations
\begin{equation}
X+y = uV, \quad Xy = u-V,
\end{equation}
to obtain
\begin{equation}
\frac{\partial X}{\partial u} = V + u \frac{\partial V}{\partial u}, \quad \frac{\partial X}{\partial y} + 1 = u \frac{\partial V}{\partial y}, \quad \frac{\partial X}{\partial u} y = 1 - \frac{\partial V}{\partial u}, \quad \frac{\partial X}{\partial y} y + X = - \frac{\partial V}{\partial y}.
\end{equation}
This is a linear system of PDEs, but it still contains $V$ and $X$, so going this way seems to be more complicated.

