Помогите срочноооооо!!!!!

Решите задачу:

$(\frac{y+1}{y-1}-\frac{y}{y+1}):\frac{3y+1}{y^2+y}=\frac{(y+1)^2-y(y-1)}{(y-1)(y+1)}\cdot\frac{y(y+1)}{3y+1}=\\\ =\frac{y^2+2y+1-y^2+y}{y-1}\cdot\frac{y}{3y+1}=\frac{3y+1}{y-1}\cdot\frac{y}{3y+1}=\frac{y}{3y+1}$

$\frac{3}{x-3}-\frac{x+15}{x^2-9}-\frac{1}{x}=\frac{3x(x+3)-x(x+15)-(x^2-9)}{x(x^2-9)}=\\\ =\frac{3x^2+9x-x^2-15x-x^2+9}{x(x-3)(x+3)}=\frac{x^2-6x+9}{x(x-3)(x+3)}=\\\ =\frac{(x-3)^2}{x(x-3)(x+3)}=\frac{x-3}{x(x+3)}$