Answer
$$dz = \left( {3{y^2} - 6{x^2}y{z^2}} \right)dx + \left( {6xy - 2{x^3}{z^2}} \right)dy - 4{x^3}yzdz$$
Work Step by Step
$$\eqalign{
& w = 3x{y^2} - 2{x^3}y{z^2} \cr
& {\text{The total differential of the dependent variable }}w{\text{ is}} \cr
& dw = \frac{{\partial w}}{{\partial x}}dx + \frac{{\partial w}}{{\partial y}}dy + \frac{{\partial w}}{{\partial z}}dz \cr
& {\text{Calculating }}\frac{{\partial w}}{{\partial x}}{\text{, }}\frac{{\partial w}}{{\partial y}}{\text{ and }}\frac{{\partial w}}{{\partial z}} \cr
& \frac{{\partial w}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {3x{y^2} - 2{x^3}y{z^2}} \right] \cr
& \frac{{\partial w}}{{\partial x}} = 3\left( 1 \right){y^2} - \left( {6{x^2}} \right)y{z^2} \cr
& \frac{{\partial w}}{{\partial x}} = 3{y^2} - 6{x^2}y{z^2} \cr
& \cr
& \frac{{\partial w}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {3x{y^2} - 2{x^3}y{z^2}} \right] \cr
& \frac{{\partial w}}{{\partial y}} = 2x\left( {2y} \right) - 2{x^3}\left( 1 \right){z^2} \cr
& \frac{{\partial w}}{{\partial y}} = 6xy - 2{x^3}{z^2} \cr
& \cr
& and \cr
& \cr
& \frac{{\partial w}}{{\partial z}} = \frac{\partial }{{\partial z}}\left[ {3x{y^2} - 2{x^3}y{z^2}} \right] \cr
& \frac{{\partial w}}{{\partial z}} = 0 - - 2{x^3}y\left( {2z} \right) \cr
& \frac{{\partial w}}{{\partial z}} = - 4{x^3}yz \cr
& \cr
& {\text{Therefore,}} \cr
& dw = \frac{{\partial w}}{{\partial x}}dx + \frac{{\partial w}}{{\partial y}}dy + \frac{{\partial w}}{{\partial z}}dz \cr
& dz = \left( {3{y^2} - 6{x^2}y{z^2}} \right)dx + \left( {6xy - 2{x^3}{z^2}} \right)dy - 4{x^3}yzdz \cr} $$
