Answer
$$dz = \frac{y}{{1 + {x^2}{y^2}}}dx + \frac{x}{{1 + {x^2}{y^2}}}dy$$
Work Step by Step
$$\eqalign{
& z = {\tan ^{ - 1}}xy \cr
& {\text{Let }}z = f\left( {x,y} \right):{\text{ }} \cr
& {\text{Calculate the partial derivative }}{f_x}\left( {x,y} \right){\text{, treating }}y{\text{ as a constant}} \cr
& {\text{ }}{f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{{\tan }^{ - 1}}xy} \right] \cr
& {\text{use }}\frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}u} \right] = \frac{1}{{{u^2} + 1}}\frac{{du}}{{dx}} \cr
& {\text{ }}{f_x}\left( {x,y} \right) = \frac{1}{{1 + {{\left( {xy} \right)}^2}}}\frac{\partial }{{\partial x}}\left[ {xy} \right] \cr
& {\text{ }}{f_x}\left( {x,y} \right) = \frac{y}{{1 + {x^2}{y^2}}} \cr
& {\text{Calculate the partial derivative }}{f_y}\left( {x,y} \right){\text{, treating }}x{\text{ as a constant}} \cr
& {\text{ }}{f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{{\tan }^{ - 1}}xy} \right] \cr
& {\text{use }}\frac{d}{{dy}}\left[ {{{\tan }^{ - 1}}u} \right] = \frac{1}{{{u^2} + 1}}\frac{{du}}{{dy}} \cr
& {\text{ }}{f_y}\left( {x,y} \right) = \frac{1}{{1 + {{\left( {xy} \right)}^2}}}\frac{\partial }{{\partial y}}\left[ {xy} \right] \cr
& {\text{ }}{f_y}\left( {x,y} \right) = \frac{x}{{1 + {x^2}{y^2}}} \cr
& \cr
& {\text{The total differential of }}z{\text{ is given by }}dz = {f_x}\left( {x,y} \right)dx + {f_y}\left( {x,y} \right)dy \cr
& {\text{Substitute the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& dz = \frac{y}{{1 + {x^2}{y^2}}}dx + \frac{x}{{1 + {x^2}{y^2}}}dy \cr} $$
