Solve: From $y=\frac{3 x^{2}-2}{x^{2}+1}$, we get $x^{2}=\frac{2+y}{3-y}$. Since $x \geqslant 0$, we have $x=\sqrt{\frac{2+y}{3-y}}$, so $f^{-1}(x)=\sqrt{\frac{2+x}{3-x}}$. From $\frac{2+x}{3-x} \geqslant 0$, we get $-2 \leqslant x<3$, so the domain of the inverse function is $[-2,3)$, hence the range of the original function is $[-2,3)$.