Answer
Vertical asymptote at $x=1$ and at $x=-1.$
Work Step by Step
$h(x)=\dfrac{x^2-9}{x^3+3x^2-x-3}=\dfrac{(x+3)(x-3)}{(x+3)(x-1)(x+1)}$
$=\dfrac{x-3}{(x-1)(x+1)}.$
Vertical asymptotes occur when the denominator alone is $0\to$
$(x-1)(x+1)=0\to x=1$ or $x=-1.$
