Answer
$${f^{ - 1}}\left( x \right) = - 2 - \sqrt {1 - x} ,{\text{ with }}x \leqslant - 2$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = - {x^2} - 4x - 3,{\text{ }}x \leqslant - 2 \cr
& y = - {x^2} - 4x - 3 \cr
& {\text{Completing the square}} \cr
& y = - \left( {{x^2} + 4x + 3} \right) \cr
& y = - \left( {{x^2} + 4x + 4 - 1} \right) \cr
& y = - \left[ {{{\left( {x + 2} \right)}^2} - 1} \right] \cr
& y = 1 - {\left( {x + 2} \right)^2} \cr
& {\text{Interchange }}x{\text{ and }}y \cr
& x = 1 - {\left( {y + 2} \right)^2} \cr
& {\text{Solve for }}y \cr
& x - 1 = - {\left( {y + 2} \right)^2} \cr
& {\left( {y + 2} \right)^2} = 1 - x \cr
& y + 2 = \pm \sqrt {1 - x} \cr
& {\text{With }}x \leqslant - 2,{\text{ so}} \cr
& y + 2 = - \sqrt {1 - x} \cr
& y = - 2 - \sqrt {1 - x} \cr
& {\text{Then}}{\text{,}} \cr
& {f^{ - 1}}\left( x \right) = - 2 - \sqrt {1 - x} ,{\text{ with }}x \leqslant - 2 \cr} $$
