Answer
Solution set: the empty set, $\emptyset$.
Work Step by Step
Follow the "Procedure for Solving Polynomial lnequalities",\ p.412:
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a polynomial function.
$x^{2}-6x+9 < 0$
We recognize the LHS as a square of a difference...
$(x-3)^{2} < 0$
...
We do not need to follow the steps any further, since this inequality asks:
"For which real numbers is a square negative?"
The answer: "For none (for no x)."
Solution set: the empty set, $\emptyset$.