Answer
$$\eqalign{
& {z_{xx}} = - {y^2}\cos xy \cr
& {z_{xy}} = - y\left( {x\cos xy} \right) - \sin xy \cr
& {z_{yy}} = - {x^2}\cos xy \cr
& {z_{yx}} = - xy\cos xy - \sin xy \cr} $$
Work Step by Step
$$\eqalign{
& z = \cos xy \cr
& {\text{Calculate the second partial derivatives }}{z_{xx}},{\text{ }}{z_{yy}},{\text{ }}{z_{xy}},{\text{ }}{x_{yx}} \cr
& {z_x} = \frac{\partial }{{\partial x}}\left[ {\cos xy} \right] \cr
& {z_x} = - y\sin xy \cr
& {z_{xy}} = \frac{\partial }{{\partial y}}\left[ { - y\sin xy} \right] \cr
& {z_{xy}} = - y\left( {x\cos xy} \right) - \sin xy \cr
& {z_{xy}} = - xy\cos xy - \sin xy \cr
& {z_{xx}} = \frac{\partial }{{\partial y}}\left[ { - y\sin xy} \right] = - {y^2}\cos xy \cr
& and \cr
& {z_y} = \frac{\partial }{{\partial y}}\left[ {\cos xy} \right] = - x\sin xy \cr
& {z_{yx}} = \frac{\partial }{{\partial y}}\left[ { - x\sin xy} \right] = - xy\cos xy - \sin xy \cr
& {z_{yy}} = \frac{\partial }{{\partial y}}\left[ { - x\sin xy} \right] = - {x^2}\cos xy \cr} $$
