Answer
$${f_{xx}}\left( {x,y} \right) = 2,\,\,\,\,\,\,\,\,{\text{ }}{f_{yy}}\left( {x,y} \right) = 18,\,\,\,\,{f_{xy}}\left( {x,y} \right) = 6{\text{ and }}{f_{yx}}\left( {x,y} \right) = 6$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {\left( {x + 3y} \right)^2} \cr
& {\text{expand using }}{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} \cr
& f\left( {x,y} \right) = {x^2} + 6xy + 9{y^2} \cr
& \cr
& {\text{Find the first partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right){\text{ then}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^2} + 6xy + 9{y^2}} \right] \cr
& {\text{treat }}y{\text{ as a constant}}{\text{, then}} \cr
& {f_x}\left( {x,y} \right) = 2x + 6y \cr
& and \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^2} + 6xy + 9{y^2}} \right] \cr
& {\text{treat }}x{\text{ as a constant}}{\text{, then }} \cr
& {f_y}\left( {x,y} \right) = 6x + 18y \cr
& \cr
& {\text{Find the second partial derivatives }}{f_{xy}}\left( {x,y} \right){\text{ and }}{f_{yx}}\left( {x,y} \right){\text{ then}} \cr
& {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2x + 6y} \right] \cr
& {f_{xy}}\left( {x,y} \right) = 6 \cr
& and \cr
& {\text{ }}{f_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {6x + 18y} \right] \cr
& {\text{ }}{f_{yx}}\left( {x,y} \right) = 6 \cr
& \cr
& {\text{Find the second partial derivatives }}{h_{xx}}\left( {x,y} \right){\text{ and }}{h_{yy}}\left( {x,y} \right){\text{ then}} \cr
& {f_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {2x + 6y} \right] \cr
& {f_{xx}}\left( {x,y} \right) = 2 \cr
& and \cr
& {\text{ }}{f_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {6x + 18y} \right] \cr
& {\text{ }}{f_{yy}}\left( {x,y} \right) = 18 \cr} $$