Sketching x³/(x² − 1)

Domain and parity. $f$ is defined on $\mathbb{R}\setminus\{\pm1\}$ and is odd: we study $x\ge0$ and use the symmetry about the origin.

Asymptotes. $x=\pm1$ are vertical asymptotes. Since $f(x)-x=\frac x{x^2-1}\to0$ at $\pm\infty$, the line $y=x$ is an oblique asymptote. The curve lies above it for $x>1$ and below it for $x<-1$ (the sign of $\frac x{x^2-1}$ is positive for $x>1$ and negative for $x<-1$).

First derivative.
$$f'(x)=\frac{x^2(x^2-3)}{(x^2-1)^2}.$$
So $f'<0$ on $(0,1)$ and $(1,\sqrt3)$, $f'>0$ for $x>\sqrt3$, and $f'(0)=0$.

Variations on $[0,\infty)$. $f$ decreases from $f(0)=0$ to $-\infty$ on $[0,1)$ (since $f\to-\infty$ as $x\to1^-$), then decreases from $+\infty$ (as $x\to1^+$) to the local minimum $f(\sqrt3)=\frac{3\sqrt3}2\approx2.6$ on $(1,\sqrt3]$, and then increases to $+\infty$, approaching the asymptote. By oddness there is a local maximum $f(-\sqrt3)=-\frac{3\sqrt3}2$.

Second derivative. $f''(x)=\dfrac{2x(x^2+3)}{(x^2-1)^3}$. It changes sign at $0$ (and at the poles $\pm1$): the origin is an inflection point where the tangent is horizontal ($f'(0)=0$), and the curve crosses it ($f(x)\approx-x^3$ near $0$).

Remark. The origin is a *stationary inflection point*: $f'(0)=0$ but $0$ is not an extremum, since $f'\le0$ on both sides of $0$ (it does not change sign there).