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