Answer
$$dz = 3{x^2}{y^2}dx + 2{x^3}ydy$$
Work Step by Step
$$\eqalign{
& z = {x^3}{y^2} \cr
& {\text{Let }}z = f\left( {x,y} \right):{\text{ }} \cr
& {\text{Calculate the partial derivative }}{f_x}\left( {x,y} \right){\text{, treating }}y{\text{ as a constant}} \cr
& {\text{ }}{f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^3}{y^2}} \right] \cr
& {\text{ }}{f_x}\left( {x,y} \right) = {y^2}\frac{\partial }{{\partial x}}\left[ {{x^3}} \right] \cr
& {\text{ }}{f_x}\left( {x,y} \right) = {y^2}\left( {3{x^2}} \right) \cr
& {\text{ }}{f_x}\left( {x,y} \right) = 3{x^2}{y^2} \cr
& {\text{Calculate the partial derivative }}{f_y}\left( {x,y} \right){\text{, treating }}x{\text{ as a constant}} \cr
& {\text{ }}{f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^3}{y^2}} \right] \cr
& {\text{ }}{f_y}\left( {x,y} \right) = {x^3}\frac{\partial }{{\partial y}}\left[ {{y^2}} \right] \cr
& {\text{ }}{f_y}\left( {x,y} \right) = {x^3}\left( {2y} \right) \cr
& {\text{ }}{f_y}\left( {x,y} \right) = 2{x^3}y \cr
& \cr
& {\text{The total differential of }}z{\text{ is given by }}dz = {f_x}\left( {x,y} \right)dx + {f_y}\left( {x,y} \right)dy \cr
& {\text{Substitute the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& dz = 3{x^2}{y^2}dx + 2{x^3}ydy \cr} $$
