Answer
$$\eqalign{
& {f_x}\left( {x,y,z} \right) = 2{z^2} + 6yz - 5{y^3} \cr
& {f_y}\left( {x,y,z} \right) = 6xz - 15x{y^2} \cr
& {f_z}\left( {x,y,z} \right) = 4xz + 6xy \cr} $$
Work Step by Step
$$\eqalign{
& f\left( {x,y,z} \right) = 2x{z^2} + 6xyz - 5x{y^3} \cr
& {\text{Calculate }}{f_x}\left( {x,y,z} \right){\text{ treating }}y{\text{ and }}z{\text{ as constants}} \cr
& {f_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {2x{z^2} + 6xyz - 5x{y^3}} \right] \cr
& {f_x}\left( {x,y,z} \right) = 2\left( 1 \right){z^2} + 6\left( 1 \right)yz - 5\left( 1 \right){y^3} \cr
& {f_x}\left( {x,y,z} \right) = 2{z^2} + 6yz - 5{y^3} \cr
& {\text{Calculate }}{f_y}\left( {x,y,z} \right){\text{ treating }}x{\text{ and }}z{\text{ as constants}} \cr
& {f_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {2x{z^2} + 6xyz - 5x{y^3}} \right] \cr
& {f_y}\left( {x,y,z} \right) = 0 + 6x\left( 1 \right)z - 5x\left( {3{y^2}} \right) \cr
& {f_y}\left( {x,y,z} \right) = 6xz - 15x{y^2} \cr
& {\text{Calculate }}{f_z}\left( {x,y,z} \right){\text{ treating }}x{\text{ and }}y{\text{ as constants}} \cr
& {f_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {2x{z^2} + 6xyz - 5x{y^3}} \right] \cr
& {f_z}\left( {x,y,z} \right) = 2x\left( {2z} \right) + 6xy\left( 1 \right) - 0 \cr
& {f_z}\left( {x,y,z} \right) = 4xz + 6xy \cr} $$
