y^2-3y
(y^2-6y+9)/(y^2-3y)
=[(y-3)(y-3)]/[y(y-3)] (on factorising the nr and the dr)
=(y-3)/y
Or (y/y - 3/y) = (1-3/y)
Note: Please refer to the notes, explanation, and other details about factorising quadratic expressions from the answer to the problem
(x^2 - 2x - 35)/(x+5) given in the site