Answer
$\dfrac{\partial F}{\partial y}=xe^{x(x+y)}$
Work Step by Step
We have $\dfrac{\partial F}{\partial u}=e^{u+v}; \dfrac{\partial F}{\partial v}=e^{u+v}; \dfrac{\partial u}{\partial y}=0; \dfrac{\partial v}{\partial y}=x$
We apply the chain rule to obtain:
$\dfrac{\partial F}{\partial y}=\dfrac{\partial F}{\partial u} \dfrac{\partial u}{\partial y}+\dfrac{\partial F}{\partial v} \dfrac{\partial v}{\partial y} \\=0+xe^{u+v}$
We are given that $u=x^2; v =xy$
Thus, $\dfrac{\partial F}{\partial y}=xe^{x^2+xy}$
Therefore, we get $\dfrac{\partial F}{\partial y}=xe^{x(x+y)}$
