We have $\dfrac {\partial u}{\partial x} = \dfrac {(1-xy)-1 (x+y)(-y)}{(1-xy)^2} = \dfrac {1+y^2}{(1-xy)^2}$

and $ \dfrac {\partial u}{\partial y} = \dfrac {(1-xy)-1 (x+y)(-x)}{(1-xy)^2} = \dfrac {1+x^2}{(1-xy)^2}$

$\dfrac {\partial v}{\partial x} = \dfrac {1}{1+x^2} , \ \dfrac {\partial v}{\partial y} = \dfrac {1}{1+y^2}$

Hence,

$\dfrac {\partial (u,v)}{\partial (x,y)} = J = \begin{vmatrix} u_x &u_y \\v_x &v_y \end{vmatrix} = \begin{vmatrix} \frac {1+y^2}{(1-xy)^2} & \frac {1+x^2}{(1-xy)^2}\\ \frac {1}{1+x^2} & \frac {1}{1+y^2} \end{vmatrix}$

$\therefore J=\dfrac {1}{(1-xy)^2} - \dfrac {1}{(1-xy)^2} = 0 $

$\therefore \dfrac {\partial (u,v)}{\partial (x,y)}=0$