Answer
$f^{-1}(x)=\frac{1}{2}sin^{-1}(\frac{x}{3})$
range of $f(x)$: $[-3,3]$
domain and range of $f^{-1}(x)$: $[-3,3]$ and $[-\frac{\pi}{4}, \frac{\pi}{4}]$
Work Step by Step
Step 1. $f(x)=3sin(2x) \longrightarrow y=3sin(2x) \longrightarrow x=3sin(2y) \longrightarrow y=\frac{1}{2}sin^{-1}(\frac{x}{3}) \longrightarrow f^{-1}(x)=\frac{1}{2}sin^{-1}(\frac{x}{3})$
Step 2. We can find the domain and range of $f(x)$: $[-\frac{\pi}{4}, \frac{\pi}{4}]$ and $[-3,3]$
Step 3. We can find the domain and range of $f^{-1}(x)$: $[-3,3]$ and $[-\frac{\pi}{4}, \frac{\pi}{4}]$