Answer
$${f_x}\left( {x,y} \right) = - xy\sin xy + \cos xy{\text{ and }}{f_y}\left( {x,y} \right) = - {x^2}\sin xy$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( {x,y} \right) = x\cos xy \cr
& {\text{Calculate }}{f_x}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {x\cos xy} \right] \cr
& {\text{using the product rule}} \cr
& {f_x}\left( {x,y} \right) = x\frac{\partial }{{\partial x}}\left[ {\cos xy} \right] + \cos xy\frac{\partial }{{\partial x}}\left[ x \right] \cr
& {f_x}\left( {x,y} \right) = x\left( { - y\sin xy} \right) + \cos xy\left( 1 \right) \cr
& {f_x}\left( {x,y} \right) = - xy\sin xy + \cos xy \cr
& \cr
& {\text{Calculate }}{f_y}\left( {x,y} \right) \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {x\cos xy} \right] \cr
& {f_y}\left( {x,y} \right) = x\frac{\partial }{{\partial y}}\left[ {\cos xy} \right] \cr
& {f_y}\left( {x,y} \right) = x\left( { - x\sin xy} \right) \cr
& {f_y}\left( {x,y} \right) = - {x^2}\sin xy \cr
& \cr
& {f_x}\left( {x,y} \right) = - xy\sin xy + \cos xy{\text{ and }}{f_y}\left( {x,y} \right) = - {x^2}\sin xy \cr} $$
