First of all, we can observe that
[tex]x^2-6x+9 = (x-3)^2[/tex]
So the expression becomes
[tex]\dfrac{x^2-4}{\sqrt{(x-3)^2}} = \dfrac{x^2-4}{|x-3|}[/tex]
This means that the expression is defined for every [tex]x\neq 3[/tex]
Now, since the denominator is always positive (when it exists), the fraction can only be positive if the denominator is also positive: we must ask
[tex]x^2-4 \geq 0 \iff x\leq -2 \lor x\geq 2[/tex]
Since we can't accept 3 as an answer, the actual solution set is
[tex](-\infty,-2] \cup [2,3) \cup (3,\infty)[/tex]
