$\frac{4y+24}{5y^2-45}+\frac{y+3}{5y^2-15y}=\frac{y-3}{y^2+3y}$
$\frac{4y+24}{5(y^2-9)}+\frac{y+3}{5y(y-3)}-\frac{y-3}{y(y+3)}=0$
$\frac{4y+24}{5(y-3)(y+3)}+\frac{y+3}{5y(y-3)}-\frac{y-3}{y(y+3)}=0$
$\frac{y(4y+24)+(y+3{)}^2-5(y-3{)}^2}{5y(y-3)(y+3)}=0$
$\frac{4y^2+24y+y^2+6y+9-5(y^2-6y+9)}{5y(y-3)(y+3)}=0$
$\frac{4y^2+24y+y^2+6y+9-5y^2+30y-45}{5y(y-3)(y+3)}=0$
$\frac{60y-36}{5y(y-3)(y+3)}=0$
$\{\begin{array}{ll}5y\ne 0& \\ y-3\ne 0& \\ y+3\ne 0& \\ 60y-36=0& \end{array}$
$\{\begin{array}{ll}y\ne 0& \\ y\ne 3& \\ y\ne -3& \\ 60y=36& \end{array}$
$\{\begin{array}{ll}y\ne 0& \\ y\ne 3& \\ y\ne -3& \\ y=\frac{36}{60}& \end{array}$
$\{\begin{array}{ll}y\ne 0& \\ y\ne 3& \\ y\ne -3& \\ y=\frac{3}{5}& \end{array}$
Ответ: $y=\frac{3}{5}$

$\frac{y+2}{8y^3+1}-\frac{1}{4y+2}=\frac{y+3}{8y^2-4y+2}$
$\frac{y+2}{8y^3+1}-\frac{1}{4y+2}-\frac{y+3}{8y^2-4y+2}=0$
$\frac{y+2}{(2y+1)(4y^2-2y+1)}-\frac{1}{2(2y+1)}-\frac{y+3}{2(4y^2-2y+1)}=0$
$\frac{2(y+2)-(4y^2-2y+1)-(2y+1)(y+3)}{2(2y+1)(4y^2-2y+1)}=0$
$\frac{2y+4-4y^2+2y-1-(2y^2+y+6y+3)}{2(2y+1)(4y^2-2y+1)}=0$
$\frac{4y+3-4y^2-2y^2-y-6y-3}{2(2y+1)(4y^2-2y+1)}=0$
$\frac{-6y^2-3y}{2(2y+1)(4y^2-2y+1)}=0$
$\frac{-3y(2y+1)}{2(2y+1)(4y^2-2y+1)}=0$
$\frac{-3y}{2(4y^2-2y+1)}=0$
$\{\begin{array}{ll}2(4y^2-2y+1)\ne 0& \\ -3y=0& \end{array}$
$\{\begin{array}{ll}4y^2-2y+1\ne 0& \\ y=0& \end{array}$
Ответ: y = 0