The domain of the function $f(x)=\frac{{\sin }^{-1}(x-3)}{\sqrt{9-x^2}}$ is

To find the domain of $f(x),$ two conditions must be met:For the numerator ${\sin }^{-1}(x-3),$ the argument must be in $[-1,1]:$$-1\le x-3\le 1\Rightarrow 2\le x\le 4$.For the denominator $\sqrt{9-x^2}$ to be defined and non-zero, the expression inside must be strictly positive:$9-x^2>0\Rightarrow x^2<9\Rightarrow -3<x<3$.The intersection of these two conditions is:$x\in [2,4]\cap (-3,3)=[2,3)$.
Thus, the domain is $[2,3)$.