Answer
$$\eqalign{
& {\bf{a}})\,\,\,9{x^2}{y^2},\,\,\,\,\,{\bf{b}})\,\,\,6{x^3}y,\,\,{\bf{c}})\,\,\,9{y^2} \cr
& {\bf{d}})\,\,\,9{x^2},\,\,\,\,\,\,\,\,\,{\bf{e}})\,\,\,6y,\,\,\,\,\,{\bf{f}})\,\,6{x^3},\, \cr
& {\bf{g}})\,\,\,\,36,\,\,\,\,\,\,\,\,\,\,{\bf{h}})\,\,12 \cr} $$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( {x,y} \right) = 3{x^3}{y^2} \cr
& {\bf{a}})\,\,{\text{Calculate the partial derivative }}{f_x}\left( {x,y} \right),{\text{ consider }}y{\text{ as a constant}} \cr
& \,\,\,\,{f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {3{x^3}{y^2}} \right] \cr
& \,\,\,\,{f_x}\left( {x,y} \right) = 3\left( {3{x^2}} \right){y^2} \cr
& \,\,\,\,{f_x}\left( {x,y} \right) = 9{x^2}{y^2} \cr
& \cr
& {\bf{b}})\,\,{\text{Calculate the partial derivative }}{f_y}\left( {x,y} \right),{\text{ consider }}x{\text{ as a constant}} \cr
& \,\,\,\,{f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {3{x^3}{y^2}} \right] \cr
& \,\,\,\,{f_y}\left( {x,y} \right) = 3{x^3}\left( {2y} \right) \cr
& \,\,\,\,{f_y}\left( {x,y} \right) = 6{x^3}y \cr
& \cr
& {\bf{c}})\,\,{\text{Find the partial derivative }}{f_x}\left( {1,y} \right),{\text{ }} \cr
& \,\,\,\,{f_x}\left( {x,y} \right) = 9{x^2}{y^2} \cr
& \,\,\,\,{f_x}\left( {1,y} \right) = 9{\left( 1 \right)^2}{\left( y \right)^2} \cr
& \,\,\,\,{f_x}\left( {1,y} \right) = 9{y^2} \cr
& \cr
& {\bf{d}})\,\,{\text{Find the partial derivative }}{f_x}\left( {x,1} \right),{\text{ }} \cr
& \,\,\,\,{f_x}\left( {x,1} \right) = 9{x^2}{y^2} \cr
& \,\,\,\,{f_x}\left( {x,1} \right) = 9{\left( x \right)^2}{\left( 1 \right)^2} \cr
& \,\,\,\,{f_x}\left( {x,1} \right) = 9{x^2} \cr
& \cr
& {\bf{e}})\,\,{\text{Find the partial derivative }}{f_y}\left( {1,y} \right),{\text{ }} \cr
& \,\,\,\,{f_y}\left( {x,y} \right) = 6{x^3}y \cr
& \,\,\,\,{f_y}\left( {1,y} \right) = 6{\left( 1 \right)^3}\left( y \right) \cr
& \,\,\,\,{f_y}\left( {1,y} \right) = 6y \cr
& \cr
& {\bf{f}})\,\,{\text{Find the partial derivative }}{f_y}\left( {x,1} \right),{\text{ }} \cr
& \,\,\,\,{f_y}\left( {x,1} \right) = 6{x^3}y \cr
& \,\,\,\,{f_y}\left( {x,1} \right) = 6{\left( x \right)^3}\left( 1 \right) \cr
& \,\,\,\,{f_y}\left( {1,y} \right) = 6{x^3} \cr
& \cr
& {\bf{g}})\,\,{\text{Find the partial derivative }}{f_x}\left( {1,2} \right),{\text{ }} \cr
& \,\,\,\,{f_x}\left( {1,2} \right) = 9{\left( 1 \right)^2}{\left( 2 \right)^2} \cr
& \,\,\,\,{f_x}\left( {1,2} \right) = 9\left( 1 \right)\left( 4 \right) \cr
& \,\,\,\,{f_x}\left( {1,2} \right) = 36 \cr
& \cr
& {\bf{h}})\,\,{\text{Find the partial derivative }}{f_y}\left( {1,2} \right),{\text{ }} \cr
& \,\,\,\,{f_y}\left( {1,2} \right) = 6{\left( 1 \right)^3}\left( 2 \right) \cr
& \,\,\,\,{f_y}\left( {1,2} \right) = 12 \cr} $$