Answer
$\displaystyle \frac{3}{x-3}, \qquad x\neq 3$
Work Step by Step
Numerator, factored:$\qquad$ $3x-9=3(x-3)$
Denominator, factored:
$x^{2}-6x+9=x^{2}-2\cdot x\cdot 3+3^{2}$=$\qquad$ ...recognize a square of a sum
$=(x-3)^{2}=(x-3)(x-3)$
Numbers to be excluded from the domain are numbers that yield 0 in the denominator:
$x\neq 3$
Expression = $\displaystyle \frac{3(x-3)}{(x-3)(x-3)}$=$\qquad$... cancel $(x-3)$
= $\displaystyle \frac{3}{x-3}, \qquad x\neq 3$
