Answer
$(1-pq)e^{p+q}$
Work Step by Step
Here, we have $\dfrac{\partial x}{\partial p}=e^q$ and $\dfrac{\partial x}{\partial q}=p e^q$
Also, $\dfrac{\partial y}{\partial p}=q e^p$ and $\dfrac{\partial y}{\partial q}= e^p$
Now, $Jacobian =\begin{vmatrix} \dfrac{\partial x}{\partial p}&\dfrac{\partial x}{\partial q}\\\dfrac{\partial y}{\partial p}&\dfrac{\partial y}{\partial q}\end{vmatrix}=\begin{vmatrix} e^q&pe^q\\ qe^p&e^p\end{vmatrix}=e^{p+q}-pqe^{p+q}=(1-pq)e^{p+q}$
