Answer
$\dfrac{\partial^2 f}{\partial x^2}=0$
$\dfrac{\partial^2 f}{\partial y^2}=0$
$\dfrac{\partial^2 f}{\partial x\partial y}=\dfrac{\partial^2 f}{\partial y \partial x }=1$
Work Step by Step
Our aim is to take the first partial derivative of the given function $f(x,y)$ with respect to one of its variables, by treating all others as a constant.
$f_x= \dfrac{\partial (x+y+xy)}{\partial x}=1+y$
$f_y=\dfrac{\partial (x+y+xy)}{\partial y}=1+x$
$\dfrac{\partial^2 f}{\partial x^2}=\dfrac{\partial }{\partial x} (1+y)=0$
$\dfrac{\partial^2 f}{\partial y^2}=\dfrac{\partial } {\partial y} (1+x)=0$
Next,
$\dfrac{\partial^2 f}{\partial x\partial y}=\dfrac{\partial^2 f}{\partial y \partial x }=\dfrac{\partial }{\partial y } (1+y)=1$
