You look good, except for one tiny little oversight.

It should be:

\(\displaystyle \L\\\frac{-(y^{2}+3y+9)}{x+1}\)

The negative comes from factoring \(\displaystyle 1-x^{2}\)

\(\displaystyle 1-x^{2}=(1-x)(x+1)\) or \(\displaystyle {-}(x-1)(x+1)\)

I assume you factored the variuos components.

\(\displaystyle \L\\\frac{\sout{(x-1)}\sout{(2y-1)}}{\sout{(y-3)}\sout{(2y-1)}}\cdot\frac{\sout{(y-3)}(y^{2}+3y+9)}{-\sout{(x-1)}(x+1)}\)

See that little negative in the denominator of the right side at the x-1.