Solution
= Solution
{c}
The <partial derivatives> are
$$
u_x=\frac{1+y^2}{(1-xy)^2},\quad u_y=\frac{1+x^2}{(1-xy)^2},\quad
v_x=\frac1{1+x^2},\quad v_y=\frac1{1+y^2}.
$$
Thus the <Jacobian determinant> is
$$
\boxed{\frac{\partial(u,v)}{\partial(x,y)}=u_xv_y-u_yv_x=0.}
$$
This vanishing reflects the functional dependence found next.