Answer
$$D = \left\{ {\left( {x,y} \right):{x^2} - 1 \leqslant y \leqslant {x^2} + 1} \right\}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {\sin ^{ - 1}}\left( {y - {x^2}} \right) \cr
& {\text{Recall that the domain of }}{\sin ^{ - 1}}\theta {\text{ is }} - 1 \leqslant \theta \leqslant 1.{\text{ Then the}} \cr
& {\text{domain of }}f\left( {x,y} \right){\text{consists of ordered pairs }}\left( {x,y} \right){\text{ for which }} \cr
& - 1 \leqslant y - {x^2} \leqslant 1 \cr
& or \cr
& {x^2} - 1 \leqslant y \leqslant {x^2} + 1 \cr
& {\text{Therefore, the domain of }}f{\text{ is }}D = \left\{ {\left( {x,y} \right):{x^2} - 1 \leqslant y \leqslant {x^2} + 1} \right\} \cr} $$
