Answer
$h_{xx}=0 \\h_{yy}=xe^y$
and $h_{xy}=h_{yx}=e^y$
Work Step by Step
Our aim is to take the first partial derivative of the given function $h(x,y)$ with respect to $x$, by treating $y$ as a constant and vice versa:
$h_x=e^y \implies h_y=xe^y+1$
Now, compute second partial derivatives:
$h_{xx}=0 \\h_{yy}=xe^y$
and $h_{xy}=h_{yx}=e^y$
