Ответ:

Ограничения: \(y\ne3\), \(y\ne-3\).

\[\begin{aligned}\frac{6}{y^2-9}+\frac{1}{3-y}&=\frac{6}{(y-3)(y+3)}-\frac{1}{y-3}\\&=\frac{6-(y+3)}{(y-3)(y+3)}\\&=-\frac1{y+3}. \end{aligned}\]

Тогда

\[ -\frac1{y+3}\cdot\frac{(y+3)^2}{5}=\frac{y+3}{5}.\]

Ответ: \(\frac{y+3}{5}\), \(y\ne\pm3\).