1. We need the domain of

$$\sin^{-1}\left[\log_3\left(\frac{x}{3}\right)\right].$$

2. For $\sin^{-1}(u)$ to be defined, its input must satisfy

$$-1 \le u \le 1.$$

So here we need

$$-1 \le \log_3\left(\frac{x}{3}\right) \le 1.$$

3. Also, the logarithm itself must be defined:

$$\frac{x}{3} > 0 \quad \Rightarrow \quad x>0.$$

But this will automatically be satisfied by the inequality solution below.

4. Now solve

$$-1 \le \log_3\left(\frac{x}{3}\right) \le 1.$$

Since base $3>1$, $\log_3 y$ is increasing, so we can convert directly to exponential form:

$$3^{-1} \le \frac{x}{3} \le 3^1.$$

That is,

$$\frac{1}{3} \le \frac{x}{3} \le 3.$$

Multiplying throughout by $3$,

$$1 \le x \le 9.$$

5. Hence the domain is

$$[1,9].$$

6. Checking options:

• A: $[1,9]$ ✅
• B: $[-1,9]$ ❌ includes negative values, invalid for logarithm
• C: $[9,1]$ ❌ reversed interval
• D: $[-9,-1]$ ❌ invalid for logarithm

Therefore, the correct answer is A.