How prove exist continous g(x) such f(g(x))=x for $f(x)=\frac{2x^3-3}{3(x^2-1)}$? Let $f(x)=\frac{2x^3-3}{3(x^2-1)}$ for $x\neq {1,-1}$. Prove exist continous g(x) satisfy f(g(x))=x and g(x)>x with every x is belong to R.

$\frac{2g(x)^3-3}{3(g(x)^2-1)}=x$ but how solve? Maybe another way?
Plotting $f$ one sees that its graph consists of three pieces. In particular $f$ is monotonically increasing on the interval $J:=\ ]1,\infty[\ $, and one has
$$\lim_{x\to1+}f(x)=-\infty,\qquad\lim_{x\to\infty} f(x)=\infty\ .$$



Therefore the restriction $f\restriction J$ maps $J$ homeomorphically onto ${\mathbb R}$, and there is a continuous inverse $g:\ {\mathbb R}\to J$. In particular $(f\restriction J)\circ g={\rm id}_{\mathbb R}$, or
$$f\bigl(g(x)\bigr)=x\qquad\forall x\in{\mathbb R}\ .$$
The graph of $f\restriction J$ lies completely to the right of the line $y=x$, which implies $g(y)&gt;y$ for all $y\in{\mathbb R}$.
$f(g(x))=x$ so $g(x)$ is inverse of $f(x)$ so nothing but reflection of f(x) on y=x. proceed , you might have to check at $x=1$ though