Symmetry classification

Substitute $-x$ for $x$: $w(-x) = \dfrac{(-x)^3}{(-x)^2-9} = \dfrac{-x^3}{x^2-9}$.

This equals $-w(x)$, so the function satisfies $w(-x) = -w(x)$, meaning $w$ is odd (symmetric about the origin).

Set the denominator equal to zero: $x^2 - 9 = 0$, giving $x = 3$ and $x = -3$.

At both values the numerator $x^3$ is nonzero (27 and -27 respectively), so both are vertical asymptotes: $x=3$ and $x=-3$.

Dividing $x^3$ by $x^2 - 9$: $x^3 = x\cdot(x^2-9) + 9x$, so $w(x) = x + \dfrac{9x}{x^2-9}$.

As $x \to \pm\infty$, the remainder term $\dfrac{9x}{x^2-9} \to 0$, so the graph approaches the line $y = x$. This is the slant asymptote.