Answer
$$ - 8uv$$
Work Step by Step
$$\eqalign{
& x = {u^2} + {v^2}{\text{, }}y = {u^2} - {v^2} \cr
& {\text{Find the partial derivatives }}\frac{{\partial x}}{{\partial u}}{\text{ and }}\frac{{\partial x}}{{\partial v}} \cr
& \frac{{\partial x}}{{\partial u}} = \frac{\partial }{{\partial u}}\left[ {{u^2} + {v^2}} \right] = 2u \cr
& \frac{{\partial x}}{{\partial v}} = \frac{\partial }{{\partial v}}\left[ {{u^2} + {v^2}} \right] = 2v \cr
& {\text{Find the partial derivatives }}\frac{{\partial y}}{{\partial u}}{\text{ and }}\frac{{\partial y}}{{\partial v}} \cr
& \frac{{\partial y}}{{\partial u}} = \frac{\partial }{{\partial u}}\left[ {{u^2} - {v^2}} \right] = 2u \cr
& \frac{{\partial y}}{{\partial v}} = \frac{\partial }{{\partial v}}\left[ {{u^2} - {v^2}} \right] = - 2v \cr
& {\text{Calculate the Jacobian}} \cr
& \frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}} = \frac{{\partial x}}{{\partial u}}\frac{{\partial y}}{{\partial v}} - \frac{{\partial y}}{{\partial u}}\frac{{\partial x}}{{\partial v}}{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Substituting the partial derivatives into }}\left( {\bf{1}} \right) \cr
& \frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}} = \left( {2u} \right)\left( { - 2v} \right) - \left( {2u} \right)\left( {2v} \right) \cr
& \frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}} = - 4uv - 4uv \cr
& \frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}} = - 8uv \cr} $$
