Answer
$${f_x}\left( {x,y} \right) = {\left( {y - 2} \right)^2}{e^{x + 2y}}{\text{ and }}{f_y}\left( {x,y} \right) = 2\left( {y - 2} \right)\left( {y - 1} \right){e^{x + 2y}}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {\left( {y - 2} \right)^2}{e^{x + 2y}} \cr
& {\text{find }}{f_x}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{{\left( {y - 2} \right)}^2}{e^{x + 2y}}} \right] \cr
& {\text{treating }}y{\text{ as a constant}} \cr
& {f_x}\left( {x,y} \right) = {\left( {y - 2} \right)^2}\frac{\partial }{{\partial x}}\left[ {{e^{x + 2y}}} \right] \cr
& {\text{use }}\left( {{e^u}} \right)' = {e^u}\left( {u'} \right) \cr
& {f_x}\left( {x,y} \right) = {\left( {y - 2} \right)^2}{e^{x + 2y}}\frac{\partial }{{\partial x}}\left[ {x + 2y} \right] \cr
& {f_x}\left( {x,y} \right) = {\left( {y - 2} \right)^2}{e^{x + 2y}} \cr
& \cr
& {\text{find }}{f_y}\left( {x,y} \right) \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{{\left( {y - 2} \right)}^2}{e^{x + 2y}}} \right] \cr
& {\text{treating }}x{\text{ as a constant}}{\text{, then use the product rule}} \cr
& {f_y}\left( {x,y} \right) = {\left( {y - 2} \right)^2}\frac{\partial }{{\partial y}}\left[ {{e^{x + 2y}}} \right] + {e^{x + 2y}}\frac{\partial }{{\partial y}}\left[ {{{\left( {y - 2} \right)}^2}} \right] \cr
& {\text{solve the derivatives}} \cr
& {f_y}\left( {x,y} \right) = {\left( {y - 2} \right)^2}\left( {2{e^{x + 2y}}} \right) + {e^{x + 2y}}\left( 2 \right)\left( {y - 2} \right) \cr
& {\text{simplifying}} \cr
& {f_y}\left( {x,y} \right) = 2{\left( {y - 2} \right)^2}{e^{x + 2y}} + 2\left( {y - 2} \right){e^{x + 2y}} \cr
& {\text{factoring out }}2\left( {y - 2} \right){e^{x + 2y}} \cr
& {f_y}\left( {x,y} \right) = 2\left( {y - 2} \right){e^{x + 2y}}\left( {y - 2 + 1} \right) \cr
& {f_y}\left( {x,y} \right) = 2\left( {y - 2} \right)\left( {y - 1} \right){e^{x + 2y}} \cr} $$
