Answer
$f(x)$ has two vertical asymptotes (non-removable discontinuity) at $x=2$ and $x=-2.$
Work Step by Step
$f(x)=\dfrac{x}{x^2-4}=\dfrac{x}{(x-2)(x+2)}\to x\ne2$ and $x\ne-2$
Hence since the denominator is zero at both values, the function has non-removable discontinuities at $x=2$ and $x=-2.$
