the domain restrictions in this case will arise from the denominator being zero for a couple values of x.

\(\dfrac{x^2+5x+6}{x^2-9} = \dfrac{x^2+5x+6}{(x-3)(x+3)} \\ \text{and it is seen that the denominator will be zero at }x=\pm 3\\ \text{thus }x \neq 3,~x \neq -3\)