Answer
$${f_{xx}}\left( {x,y} \right) = 30xy{\text{ and }}{f_{xy}}\left( {x,y} \right) = 15{x^2} - 12y$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 5{x^3}y - 6x{y^2} \cr
& {\text{find }}{f_x}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {5{x^3}y - 6x{y^2}} \right] \cr
& {\text{treat y as a constant and }}x{\text{ as a variable}}{\text{. then}} \cr
& {f_x}\left( {x,y} \right) = 5y\frac{\partial }{{\partial x}}\left[ {{x^3}} \right] - 6{y^2}\frac{\partial }{{\partial x}}\left[ x \right] \cr
& {\text{solving derivatives}} \cr
& {f_x}\left( {x,y} \right) = 5y\left( {3{x^2}} \right) - 6{y^2}\left( 1 \right) \cr
& {f_x}\left( {x,y} \right) = 15{x^2}y - 6{y^2} \cr
& \cr
& {\text{find }}{f_{xx}}\left( {x,y} \right) \cr
& {\text{ }}{f_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {15{x^2}y - 6{y^2}} \right] \cr
& {\text{ }}{f_{xx}}\left( {x,y} \right) = 15y\frac{\partial }{{\partial x}}\left[ {{x^2}} \right] - \frac{\partial }{{\partial x}}\left[ {6{y^2}} \right] \cr
& {\text{ }}{f_{xx}}\left( {x,y} \right) = 30xy \cr
& \cr
& {\text{find }}{f_{xy}}\left( {x,y} \right) \cr
& {\text{differentiate }}{f_x}\left( {x,y} \right){\text{ with respect to }}y \cr
& {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {15{x^2}y - 6{y^2}} \right] \cr
& {\text{treat }}x{\text{ as a constant and }}y{\text{ as a variable}}{\text{. then}} \cr
& {f_{xy}}\left( {x,y} \right) = 15{x^2} - 12y \cr} $$
