Answer
$\dfrac{-2u}{v}$
Work Step by Step
$Jacobian =\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}$
Now, $Jacobian =\begin{vmatrix} v&u\\\dfrac{1}{v}&\dfrac{-u}{v^2}\end{vmatrix}=\dfrac{-u}{v}-\dfrac{u}{v}=\dfrac{-2u}{v}$
