Answer
$\frac{y+3}{y-3}$.
Work Step by Step
The product of a binomial sum and a binomial difference is: $(A+B)(A-B)=A^2-B^2$.
The square of a binomial difference is: $(A-B)^2=A^2-2AB+B^2$.
Hence here: $\frac{y^2-9}{y^2-6y+9}=\frac{y^2-3^2}{y^2-2\cdot y\cdot3+3^2}=\frac{(y+3)(y-3)}{(y-3)^2}=\frac{y+3}{y-3}$.