Answer
$$\dfrac{\partial^2 f}{\partial x^2}=-y^2 \sin (xy) \\\dfrac{\partial^2 f}{\partial y^2}=-x^2 \sin (xy) \\ \dfrac{\partial^2 f}{\partial x\partial y}=\dfrac{\partial^2 f}{\partial y \partial x }=-xy \sin xy+\cos xy$$
Work Step by Step
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= y \cos (xy) $
We repeat the process for the other partial derivative.
$f_y=x \cos (xy) $
Now, compute the second partial derivatives:
$\dfrac{\partial^2 f}{\partial x^2}=\dfrac{\partial }{\partial x} (y \cos (xy)=-y^2 \sin xy$
$\dfrac{\partial^2 f}{\partial y^2}=\dfrac{\partial }{\partial y} [x \cos (xy)]=-x^2 \sin xy$
Next, $\dfrac{\partial^2 f}{\partial x\partial y}=\dfrac{\partial^2 f}{\partial y \partial x }=\dfrac{\partial }{\partial y } [y \cos (xy)])=-x \space y \sin x \space y+\cos x \space y$
