jmc

We can factor the quadratics in the numerators on the left side, and lo and behold, we can simplify the fractions: $$\begin{array}{rl}\frac{y^2-9y+8}{y-1}+\frac{3y^2+16y-12}{3y-2}& =\frac{\cancel{(y-1)}(y-8)}{\cancel{y-1}}+\frac{\cancel{(3y-2)}(y+6)}{\cancel{3y-2}}\\ & =y-8+y+6.\end{array}$$So, our equation is $2y-2=-3$, which gives $y=\boxed{-\frac{1}{2}}$. (A quick check shows that this solution is not extraneous.)