Answer
$-\cos (2\theta)$
Work Step by Step
$Jacobian =\begin{vmatrix} \dfrac{\partial x}{\partial r}&\dfrac{\partial x}{\partial \theta}\\\dfrac{\partial y}{\partial r}&\dfrac{\partial \theta}{\partial v}\end{vmatrix}$
Now, $Jacobian =\begin{vmatrix} -e^r \sin \theta&-e^{-r} \cos \theta\\e^r \cos \theta&-e^r \sin \theta\end{vmatrix}=\sin^2 \theta-\cos^2 \theta=-\cos (2\theta)$