Answer
$f(x)$ has two irremovable discontinuities (vertical asymptotes) at $x=3$ and $x=-3.$
Work Step by Step
$f(x)=\dfrac{1}{x^2-9}=\dfrac{1}{(x-3)(x+3)};x\ne3, x\ne-3$
When $x=3$ or $x=-3,$ the denominator is $0$, indicating the presence of a vertical asymptote at both points.
These discontinuities are not removable.
