Answer
$$dz = {e^{xy}}\left( {ydx + xdy} \right)$$
Work Step by Step
$$\eqalign{
& z = {e^{xy}} \cr
& {\text{Let }}z = f\left( {x,y} \right):{\text{ }} \cr
& {\text{Calculate the partial derivative }}{f_x}\left( {x,y} \right){\text{, treat }}y{\text{ as a constant}} \cr
& {\text{ }}{f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{e^{xy}}} \right] \cr
& {\text{ }}{f_x}\left( {x,y} \right) = {e^{xy}}\frac{\partial }{{\partial x}}\left[ {xy} \right] \cr
& {\text{ }}{f_x}\left( {x,y} \right) = y{e^{xy}} \cr
& {\text{Calculate the partial derivative }}{f_y}\left( {x,y} \right){\text{, treat }}x{\text{ as a constant}} \cr
& {\text{ }}{f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{e^{xy}}} \right] \cr
& {\text{ }}{f_y}\left( {x,y} \right) = {e^{xy}}\frac{\partial }{{\partial y}}\left[ {xy} \right] \cr
& {\text{ }}{f_y}\left( {x,y} \right) = x{e^{xy}} \cr
& \cr
& {\text{The total differential of }}z{\text{ is given by }}dz = {f_x}\left( {x,y} \right)dx + {f_y}\left( {x,y} \right)dy \cr
& {\text{Substitute the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& dz = y{e^{xy}}dx + x{e^{xy}}dy \cr
& dz = {e^{xy}}\left( {ydx + xdy} \right) \cr} $$
