Answer
$${f_x} = 2x + y,{\text{ }}{f_y} = x,{\text{ and }}{f_{yy}} = 0$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {x^2} + xy \cr
& {\text{Find }}{f_x} \cr
& {f_x} = {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^2} + xy} \right] \cr
& {f_x} = \frac{\partial }{{\partial x}}\left[ {{x^2} + xy} \right] \cr
& {f_x} = \frac{\partial }{{\partial x}}\left[ {{x^2}} \right] + \frac{\partial }{{\partial x}}\left[ {xy} \right] \cr
& {\text{Consider }}y{\text{ as a constant}} \cr
& {f_x} = \frac{\partial }{{\partial x}}\left[ {{x^2}} \right] + y\frac{\partial }{{\partial x}}\left[ x \right] \cr
& {f_x} = 2x + y \cr
& \cr
& {\text{Find }}{f_y} \cr
& {f_y} = {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^2} + xy} \right] \cr
& {f_y} = \frac{\partial }{{\partial y}}\left[ {{x^2} + xy} \right] \cr
& {f_y} = \frac{\partial }{{\partial y}}\left[ {{x^2}} \right] + \frac{\partial }{{\partial y}}\left[ {xy} \right] \cr
& {\text{Consider }}x{\text{ as a constant}} \cr
& {f_y} = \frac{\partial }{{\partial y}}\left[ {{x^2}} \right] + x\frac{\partial }{{\partial y}}\left[ y \right] \cr
& {f_y} = 0 + x\left( 1 \right) \cr
& {f_y} = x \cr
& \cr
& {\text{Find }}{f_{yy}} \cr
& {f_{yy}} = \frac{\partial }{{\partial y}}\left[ {{f_y}} \right] \cr
& {f_{yy}} = \frac{\partial }{{\partial y}}\left[ x \right] \cr
& {\text{Consider }}x{\text{ as a constant}} \cr
& {f_{yy}} = 0 \cr
& \cr
& {f_x} = 2x + y,{\text{ }}{f_y} = x,{\text{ and }}{f_{yy}} = 0 \cr} $$
