Answer
$f(x)$ has a removable discontinuity at $x=-2$ and a non-removable discontinuity at $x=3.$
Work Step by Step
$f(x)=\dfrac{x+2}{x^2-x-6}=\dfrac{(x+2)}{(x+2)(x-3)}=\dfrac{1}{x-3}\to x\ne-2$ and $x\ne3.$
Hence, $f(x)$ has a removable discontinuity at $x=-2$ and an irremovable discontinuity (vertical asymptote) at $x=3.$
