\(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3}=\frac{3\cdot\left(1-4x^2\right)}{\left(x^2+4x\right)\cdot\left(2-4x\right)}\)
\(=\frac{3\cdot\left(1-2x\right)\cdot\left(1+2x\right)}{x\cdot\left(x+2\right)\cdot2\cdot\left(1-2x\right)}=\frac{3\cdot\left(1+2x\right)}{2x\cdot\left(x+2\right)}\)
= \(\frac{3+6x}{2x^2+4x}\)