Jacobian of implicit functions 
Question Statement:-
  If $u^3+v^3=x+y$ and $u^2+v^2=x^3+y^3$, show that $$\frac{\partial(u,v)}{\partial(x,y)}=\frac{1}{2}\frac{y^2-x^2}{uv(u-v)}$$


My Solution:-
As the relation that are given between $u,v,x \;\& \; y$ do not seemed to me to be separated so as to get the $u$ and $v$ in the terms of $x\;\&\;y$, so I went with the following approach.
We are given with the following relations:-
$$u^3+v^3=x+y\tag{1}$$
$u^2+v^2=x^3+y^3\tag{2}$
On partially differentiating the above given relations we have,
$$3u^2\frac{\partial u}{\partial{x}}+3v^2\frac{\partial{v}}{\partial{x}}=1\tag{3}$$
$$2u\frac{\partial u}{\partial{x}}+2v\frac{\partial{v}}{\partial{x}}=3x^2\tag{4}$$
The equations $(3)$ and $(4)$ can be written as a matrix as follows
$$\begin{bmatrix}
3u^2 & 3v^2\\
2u & 2v
\end{bmatrix}
\begin{bmatrix}
\dfrac{\partial{u}}{\partial{x}}\\
\dfrac{\partial{v}}{\partial{x}}
\end{bmatrix}
=
\begin{bmatrix}
1 \\ 3x^2
\end{bmatrix} \\
\implies 
\begin{bmatrix}
\dfrac{\partial{u}}{\partial{x}}\\
\dfrac{\partial{v}}{\partial{x}}
\end{bmatrix}=
\dfrac{1}{6uv(u-v)}\begin{bmatrix}
2v & -2u\\
-3v^2 & 3u^2
\end{bmatrix}
\begin{bmatrix}
1 \\ 3x^2
\end{bmatrix}\\
\implies 
\begin{bmatrix}
\dfrac{\partial{u}}{\partial{x}}\\
\dfrac{\partial{v}}{\partial{x}}
\end{bmatrix}=
\dfrac{1}{6uv(u-v)}\begin{bmatrix}
2v-6ux^2\\
9u^2x^2-3v^2
\end{bmatrix}
$$
So, we get
$$\dfrac{\partial{u}}{\partial{x}}=\dfrac{2v-6ux^2}{6uv(u-v)}\\
\&\\
\dfrac{\partial{v}}{\partial{x}}=\dfrac{9u^2x^2-3v^2}{6uv(u-v)}$$
And due to the symmetry in the equations $(1)$ and $(2)$, we get 
$$\dfrac{\partial{u}}{\partial{y}}=\dfrac{2v-6uy^2}{6uv(u-v)}\\
\&\\
\dfrac{\partial{v}}{\partial{y}}=\dfrac{9u^2y^2-3v^2}{6uv(u-v)}$$
So, we get the Jacobian determinant as 
$$\frac{\partial(u,v)}{\partial(x,y)}=
\begin{vmatrix}
\dfrac{\partial{u}}{\partial{x}} & \dfrac{\partial{u}}{\partial{y}}\\
\dfrac{\partial{v}}{\partial{x}} & \dfrac{\partial{v}}{\partial{y}}
\end{vmatrix}=
\begin{vmatrix}
\dfrac{2v-6ux^2}{6uv(u-v)} & \dfrac{2v-6uy^2}{6uv(u-v)}\\
\dfrac{9u^2x^2-3v^2}{6uv(u-v)} & \dfrac{9u^2y^2-3v^2}{6uv(u-v)}
\end{vmatrix}=\frac{1}{2}\frac{y^2-x^2}{uv(u-v)}$$

My deal with the question:-
This method seemed a bit too long, can you suggest a shorter method. If the method teaches me something new that would be good.
 A: let $f_{1}=u^{3}+v^{3}-x-y$
and $f_{2}=u^{2}+v^{2}-x^3-y^3$
we know, by property of jacobians
$\dfrac{\partial(u,v)}{\partial(x,y)}=(-1)^2\dfrac{\dfrac{\partial(f_{1},f_{2})}{\partial(x,y)}}{\dfrac{\partial(f_{1},f_{2})}{\partial(u,v)}}\tag{1}$ 
${\dfrac{\partial(f_{1},f_{2})}{\partial(x,y)}}=3(y^2-x^2)\tag{2}$
${\dfrac{\partial(f_{1},f_{2})}{\partial(u,v)}}=6\ u\ v\ (u-v)\tag{3}$
put equation $(2)$ and $(3)$  in $(1)$ ...you will have your result
A: This answer blatantly steals the comment by user @amd. Take your first matrix equation and add a column for the derivatives with respect to $y$:
$$\begin{bmatrix}
3u^2 & 3v^2\\
2u & 2v
\end{bmatrix}
\begin{bmatrix}
\dfrac{\partial{u}}{\partial{x}} & \dfrac{\partial{u}}{\partial{y}}\\
\dfrac{\partial{v}}{\partial{x}} & \dfrac{\partial{v}}{\partial{y}}
\end{bmatrix}
=
\begin{bmatrix}
1 & 1 \\ 3x^2  & 3y^2
\end{bmatrix}. \\
$$
Now we can take the determinant of both sides:
$$
(6u^2v - 6v^2u)\frac{\partial(u,v)}{\partial(x,y)} = (3y^2 - 3x^2)
$$
which implies the result.
