Answer
The set of all real numbers except $-3$ and $-1$.
Work Step by Step
The denominator cannot be equal to zero. Solving for the values that make the denominator zero in the given function, $
f(x)=\dfrac{9x^2-9}{x^2+4x+3}
,$ then
\begin{array}{l}\require{cancel}
x^2+4x+3=0
\\\\
(x+3)(x+1)=0
\\\\
x-\{-3,-1\}
.\end{array}
Hence, the domain is $\text{
the set of all real numbers except $-3$ and $-1$
.}$
