Answer
$\dfrac{\partial^2 f}{\partial x^2}=0$
$\dfrac{\partial^2 f}{\partial y^2}=0$
and
$\dfrac{\partial^2 f}{\partial x\partial y}=\dfrac{\partial^2 f}{\partial y \partial x }=1$
Work Step by Step
We need to take the first partial derivatives of the given function.
In order to find the partial derivative, we will differentiate with respect to $x$, by keeping $y$ and $z$ as a constant, and vice versa:
$f_x= \dfrac{\partial (x+y+xy)}{\partial x}=1+y$
$f_y=\dfrac{\partial (x+y+xy)}{\partial y}=1+x$
$\dfrac{\partial^2 f}{\partial x^2}=\dfrac{\partial (1+y)}{\partial x}=0$
$\dfrac{\partial^2 f}{\partial y^2}=\dfrac{\partial (1+x)}{\partial y}=0$
Now, $\dfrac{\partial^2 f}{\partial x\partial y}=\dfrac{\partial^2 f}{\partial y \partial x }=\dfrac{\partial (1+y) }{\partial y } =1$
