$$\frac{1}{3x+y}+\frac{1}{3x-4}=\frac{3}{4}\cdot\frac{1}{2(3x+4)}\frac{1}{2(8x-y)}=\frac{1}{8}$$
$\frac{6x+y-4}{\left(3x-4\right)\left(3x+y\right)}=\frac{3}{16\left(3x+4\right)\left(8x-y\right)}\text{ and }\frac{3}{16\left(3x+4\right)\left(8x-y\right)}=\frac{1}{8}$
