We define u and v as $$u=\frac{2x}{x+y+1}$$ $$v=\frac{2y}{x+y+1}$$
I'm trying to get Jacobian determinant $\frac{ \partial (x,y) }{ \partial (u,v) }$ and express it only using $u$ and $v$.
What I have tried
$$x=\frac{-uy-u}{u-2}=-\frac{2y}{v^2}$$ $$y=-\frac{2x}{u^2}=\frac{-vx-v}{v-2}$$ $$\frac{ \partial x }{ \partial u }=\frac{2y+2}{(u-2)^2}$$ $$\frac{ \partial y }{ \partial u }=-\frac{2x}{u^2}$$ $$\frac{ \partial x }{ \partial v }=\frac{2y}{v^2}$$ $$\frac{ \partial y }{ \partial v }=-\frac{2x+2}{(v-2)^2}$$
But even if you calculate the determinant with these values, it end up with result below and fail to cancel x and y out.
$$\frac{-16yu^2x+64yux-64yx+4yu^2v^2+16yuxv^2-16yxv^2+16yu^2xv-64yuxv+64yxv+4u^2xv^2+4u^2v^2}{u^2v^2\left(u-2\right)^2\left(v-2\right)^2}$$
Can anyone help me?
Calculate $\frac{\partial(u,v)}{\partial(x,y)}$ first
$$\frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} \frac{2}{x+y+1} - \frac{2x}{(x+y+1)^2} & \frac{-2x}{(x+y+1)^2} \\ \frac{-2y}{(x+y+1)^2} & \frac{2}{x+y+1}-\frac{2y}{(x+y+1)^2} \\ \end{vmatrix} = \frac{4}{(x+y+1)^3}$$
Then notice that
$$u+v = 2-\frac{2}{x+y+1} \implies x+y+1 = \frac{2}{2-u-v}$$
which means
$$\frac{\partial(x,y)}{\partial(u,v)} = \frac{(x+y+1)^3}{4} = \frac{2}{(2-u-v)^3}$$