Answer
$$\eqalign{
& \left( {\text{a}} \right)9 \cr
& \left( {\text{b}} \right)3 \cr
& \left( {\text{c}} \right)0 \cr
& \left( {\text{d}} \right)6{x^2} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 3{x^2}y \cr
& \left( {\text{a}} \right)\left( {1,3} \right) \cr
& {\text{Evaluate, substitute 1 for }}x{\text{ and 3 for }}y \cr
& f\left( {1,3} \right) = 3{\left( 1 \right)^2}\left( 3 \right) = 9 \cr
& \cr
& \left( {\text{b}} \right)\left( { - 1,1} \right) \cr
& {\text{Evaluate, substitute }} - {\text{1 for }}x{\text{ and 1 for }}y \cr
& f\left( { - 1,1} \right) = 3{\left( { - 1} \right)^2}\left( 1 \right) = 3 \cr
& \cr
& \left( {\text{c}} \right)\left( { - 4,0} \right) \cr
& {\text{Evaluate, substitute }} - 4{\text{ for }}x{\text{ and 0 for }}y \cr
& f\left( { - 4,0} \right) = 3{\left( { - 4} \right)^2}\left( 0 \right) = 0 \cr
& \cr
& \left( {\text{d}} \right)\left( {x,2} \right) \cr
& {\text{Evaluate, substitute }}2{\text{ for }}y \cr
& f\left( {x,2} \right) = 3{x^2}\left( 2 \right) = 6{x^2} \cr} $$
