Упростим выражение:
$$\left(\frac{6}{y^2-9} + \frac{1}{3-y}\right) \cdot \frac{y^2+6y+9}{5}$$
$$\left(\frac{6}{(y-3)(y+3)} - \frac{1}{y-3}\right) \cdot \frac{(y+3)^2}{5}$$
$$\left(\frac{6 - (y+3)}{(y-3)(y+3)}\right) \cdot \frac{(y+3)^2}{5}$$
$$\left(\frac{3 - y}{(y-3)(y+3)}\right) \cdot \frac{(y+3)^2}{5}$$
$$\frac{-(y-3)}{(y-3)(y+3)} \cdot \frac{(y+3)^2}{5}$$
$$-\frac{1}{(y+3)} \cdot \frac{(y+3)^2}{5}$$
$$-\frac{(y+3)}{5}$$
Ответ: $$-\frac{y+3}{5}$$.