#### Answer

$f(x)$ has a removable discontinuity at $x=0$ and two irremovable discontinuities (vertical asymptotes) at $x=1$ and $x=-1.$

#### Work Step by Step

$f(x)=\dfrac{x}{x^3-x}=\dfrac{1}{x^2-1}=\dfrac{1}{(x-1)(x+1)};x\ne-1,x\ne0,x\ne1.$
At $x=0$ the function has a removable discontinuity (since the factor cancels out from both the numerator and denominator).
At both $x=1$ and $x=-1$, the denominator is $0$, indicating the presence of a vertical asymptote at both points.
The vertical asymptotes are not removable.