Answer
(a) $ f^{-1}(x)=\frac{2x}{3-x}$.
(b) $f$: $(-\infty,-2)U(-2,\infty)$ and $(-\infty,3)U(3,\infty)$. $f^{-1}$: $(-\infty,3)U(3,\infty)$ and $(-\infty,-2)U(-2,\infty)$.
Work Step by Step
(a) $f(x)=\frac{3x}{x+2} \longrightarrow y=\frac{3x}{x+2}\longrightarrow x=\frac{3y}{y+2}\longrightarrow y=\frac{2x}{3-x}\longrightarrow f^{-1}(x)=\frac{2x}{3-x}$. Check $f(f^{-1})=\frac{3(\frac{2x}{3-x})}{(\frac{2x}{3-x})+2} =x$ and $f^{-1}(f)=\frac{2(\frac{3x}{x+2})}{3-(\frac{3x}{x+2} )}=x$
(b) Domain and range of $f$: $(-\infty,-2)U(-2,\infty)$ and $(-\infty,3)U(3,\infty)$. Domain and range of $f^{-1}$: $(-\infty,3)U(3,\infty)$ and $(-\infty,-2)U(-2,\infty)$.