Answer
$\dfrac{\partial^2 f}{\partial x^2}=-y^2 \sin (xy)$
$\dfrac{\partial^2 f}{\partial y^2}=-x^2 \sin (xy)$
and
$\dfrac{\partial^2 f}{\partial x\partial y}=\dfrac{\partial^2 f}{\partial y \partial x }=-xy \sin xy+\cos xy$
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 (\sin xy)}{\partial x}=y \cos (xy) $
$f_y=\dfrac{\partial (\sin xy)}{\partial y}=x \cos (xy) $
$\dfrac{\partial^2 f}{\partial x^2}=\dfrac{\partial (y \cos (xy)}{\partial x}=-y^2 \sin (xy)$
$\dfrac{\partial^2 f}{\partial y^2}=\dfrac{\partial [x \cos (xy)]}{\partial y}=-x^2 \sin (xy)$
Now, $\dfrac{\partial^2 f}{\partial x\partial y}=\dfrac{\partial^2 f}{\partial y \partial x }=\dfrac{\partial [y \cos (xy)]) }{\partial y } =-xy \sin xy+\cos xy$
