Answer
$f^{-1}(x)=\sqrt {\frac{3}{3x-1}}$
Work Step by Step
$f(x)=\frac{x^2+3}{3x^2},$
$y=\frac{x^2+3}{3x^2},$
a.$x=\frac{y^2+3}{3y^2},$
$3xy^2=y^2+3,$
$3xy^2-y^2=3,$
$y^2(3x-1)=3,$
$y^2=\frac{3}{3x-1},$
$y=\sqrt {\frac{3}{3x-1}}=f^{-1}(x),$
b. Domain of $f(x)$ is $x\in\mathbb{R} -\{ 0\},$ and Domain of $f^{-1}(x)$ is $x>\frac{1}{3}$
