\(y=9-x^2\) for \(x\ge3\).
Interchanging \(x\) and \(y\) yields \(x=9-y^2\) for \(y\ge3\).
Solving for \(y\) yields \(y=\pm\sqrt{9-x}\).
Since \(y\ge3\) we know that \(f^{-1}(x)=\sqrt{9-x}\) rather than \(-\sqrt{9-x}\).
Furthermore, since \(\sqrt{9-x}\ge3\) it follows that \(9-x\ge9\) which means that it must be the case that \(x\le0\).
Therefore, \(f^{-1}(x)=\sqrt{9-x}\) for \(x\le0\).