1. We need the domain of $f(x)=\frac{\sin^{-1}(x-3)}{\sqrt{9-x^2}}.$

For the function to be defined, both of these must hold:

• $\sin^{-1}(x-3)$ must be defined.
• The denominator $\sqrt{9-x^2}$ must be real and nonzero.

2. Condition from the inverse sine:

Since $\sin^{-1}(y)$ is defined for $-1\le y\le 1,$ we need $-1\le x-3\le 1.$ Adding $3$ throughout, $2\le x\le 4.$

So from the numerator, $x\in [2,4].$

3. Condition from the denominator:

The denominator is $\sqrt{9-x^2}.$ For a square root in the denominator, the expression inside must be strictly positive: $9-x^2>0.$ Thus, $x^2<9 \implies -3<x<3.$

So from the denominator, $x\in (-3,3).$

4. Intersect the two conditions:

$[2,4]\cap(-3,3)=[2,3).$

Therefore, the domain is $\boxed{[2,3)}.$

5. Check options:

• A: $[1,2]$ ❌
• B: $[2,3)$ ✅
• C: $[1,2)$ ❌
• D: $[2,3]$ ❌ since $x=3$ makes denominator zero

Hence the correct option is B.