Answer
$${f_x}\left( {x,y} \right) = 6xy + {y^3}{\text{ and }}{f_y}\left( {x,y} \right) = 3{x^2} + 3x{y^2}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( {x,y} \right) = 3{x^2}y + x{y^3} \cr
& {\text{Calculate }}{f_x}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {3{x^2}y + x{y^3}} \right] \cr
& {f_x}\left( {x,y} \right) = 6xy + {y^3} \cr
& \cr
& {\text{Calculate }}{f_y}\left( {x,y} \right) \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {3{x^2}y + x{y^3}} \right] \cr
& {f_y}\left( {x,y} \right) = 3{x^2} + 3x{y^2} \cr
& \cr
& {f_x}\left( {x,y} \right) = 6xy + {y^3}{\text{ and }}{f_y}\left( {x,y} \right) = 3{x^2} + 3x{y^2} \cr} $$