$$\frac{y^2 - 16}{4y^2 - y^3} = \frac{(y-4)(y+4)}{y^2(4-y)} = \frac{-(4-y)(y+4)}{y^2(4-y)} = -\frac{y+4}{y^2}$$
Ответ: $$\frac{-(y+4)}{y^2}$$