Решение:
$$\frac{y-2}{y^2-1} - \frac{y-1}{y^2+y} = \frac{y-2}{(y-1)(y+1)} - \frac{y-1}{y(y+1)} = \frac{y(y-2)-(y-1)^2}{y(y-1)(y+1)} = \frac{y^2-2y-(y^2-2y+1)}{y(y-1)(y+1)} = \frac{y^2-2y-y^2+2y-1}{y(y-1)(y+1)} = \frac{-1}{y(y-1)(y+1)}$$При y = -3:
$$\frac{-1}{-3(-3-1)(-3+1)} = \frac{-1}{-3(-4)(-2)} = \frac{-1}{-24} = \frac{1}{24}$$Ответ:
$$\frac{1}{24}$$