66. 4) $$\frac{21y^2+1}{1-9y^2} - \frac{y}{3y-1};$$

$$\frac{21y^2+1}{1-9y^2} - \frac{y}{3y-1} = \frac{21y^2+1}{(1-3y)(1+3y)} + \frac{y}{1-3y} = \frac{21y^2+1}{(1-3y)(1+3y)} + \frac{y(1+3y)}{(1-3y)(1+3y)} = \frac{21y^2+1+y+3y^2}{(1-3y)(1+3y)} = \frac{24y^2+y+1}{1-9y^2}$$ Ответ: $$\frac{24y^2+y+1}{1-9y^2}$$