$F = \sin (xy)$

$\dfrac{\partial F}{\partial x} = y \, \cos (xy)$

$\dfrac{\partial F}{\partial y} = x \, \cos (xy)$

$\dfrac{\partial^2 F}{\partial x \, \partial y} = \dfrac{\partial}{\partial y} \Big[ y \, \cos (xy) \Big]$

$\dfrac{\partial^2 F}{\partial x \, \partial y} = -xy \, \sin (xy) + \cos (xy)$

$\dfrac{\partial^2 F}{\partial y \, \partial x} = \dfrac{\partial}{\partial x} \Big[ x \, \cos (xy) \Big]$

$\dfrac{\partial^2 F}{\partial y \, \partial x} = -xy \, \sin (xy) + \cos (xy)$

Hence,

$\dfrac{\partial^2 F}{\partial x \, \partial y} = \dfrac{\partial^2 F}{\partial y \, \partial x}$