Answer
$$dz = \left( {xy\cos xy + \sin xy} \right)dx + \left( {{x^2}\cos xy} \right)dy$$
Work Step by Step
$$\eqalign{
& z = x\sin xy \cr
& {\text{The total differential of the dependent variable }}z{\text{ is}} \cr
& dz = \frac{{\partial z}}{{\partial x}}dx + \frac{{\partial z}}{{\partial y}}dy \cr
& {\text{Calculating }}\frac{{\partial z}}{{\partial x}}{\text{ and }}\frac{{\partial z}}{{\partial y}} \cr
& \frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {x\sin xy} \right] \cr
& \frac{{\partial z}}{{\partial x}} = x\left( {y\cos xy} \right) + \sin xy \cr
& \frac{{\partial z}}{{\partial x}} = xy\cos xy + \sin xy \cr
& and \cr
& \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {x\sin xy} \right] \cr
& \frac{{\partial z}}{{\partial y}} = {x^2}\cos xy \cr
& {\text{Therefore,}} \cr
& dz = \frac{{\partial z}}{{\partial x}}dx + \frac{{\partial z}}{{\partial y}}dy \cr
& dz = \left( {xy\cos xy + \sin xy} \right)dx + \left( {{x^2}\cos xy} \right)dy \cr} $$
