Answer
(a) $ f^{-1}(x)=\sqrt {\frac{3}{3x-1}}$.
(b) $f$: $(0,\infty)$ and $(\frac{1}{3},\infty)$.
$f^{-1}$: $(\frac{1}{3},\infty)$ and $(0,\infty)$.
Work Step by Step
(a) $f(x)=\frac{x^2+3}{3x^2}, x\gt0\longrightarrow y=\frac{x^2+3}{3x^2} \longrightarrow x=\frac{y^2+3}{3y^2}, y\gt0 \longrightarrow y=\sqrt {\frac{3}{3x-1}} \longrightarrow f^{-1}(x)=\sqrt {\frac{3}{3x-1}}$. Check $f(f^{-1})=\frac{(\sqrt {\frac{3}{3x-1}})^2+3}{3(\sqrt {\frac{3}{3x-1}})^2}=x$ and $f^{-1}(f)=\sqrt {\frac{3}{3(\frac{x^2+3}{3x^2})-1}}=x$
(b) Domain and range of $f$: $(0,\infty)$ and $(\frac{1}{3},\infty)$. Domain and range of $f^{-1}$: $(\frac{1}{3},\infty)$ and $(0,\infty)$.
