8 y² + 8y 4y - 4 - 4- y² 4-y²

$$\frac{y^2+8y}{4-y^2} - \frac{4y-4}{4-y^2} = \frac{y^2+8y-(4y-4)}{4-y^2} = \frac{y^2+8y-4y+4}{4-y^2} = \frac{y^2+4y+4}{4-y^2} = \frac{(y+2)^2}{(2-y)(2+y)} = \frac{y+2}{2-y}$$

Ответ: $$\frac{y+2}{2-y}$$