由 $z=\cos\frac{4\pi}{7}+i\sin\frac{4\pi}{7}$，知 $z$ 满足方程 $z^7-1=0$．又$$z^7-1=(z-1)(z^6+z^5+z^4+z^3+z^2+z+1),$$且 $z\neq 1$，故$$z^6+z^5+z^4+z^3+z^2+z+1=0.$$将 $\frac{z}{1+z^2}+\frac{z^2}{1+z^4}+\frac{z^3}{1+z^6}$ 通分后，分母为$$\begin{aligned}
(1+z^2)(1+z^4)(1+z^6)&=1+z^2+z^4+z^6+z^6+z^8+z^{10}+z^{12}\\
&=1+z^2+z^4+z^6+z^6+z+z^3+z^5\\
&=z^6,\\ \end{aligned}$$分子为$$\begin{aligned}
&z(1+z^4)(1+z^6)+x^2(1+z^2)(1+z^6)+z^3(1+z^2)(1+z^4)\\
&=(1+z^4)(z+1)+(1+z^2)(z^2+2)+(1+z^2)(z^3+1)\\
&=1+z+z^4+z^5+z+z^2+z^3+z^4+1+z^2+z^3+z^5\\
&=2(1+z+z^2+z^3+z^4+z^5)\\ &=-2z^6.\\
\end{aligned}$$从而$$\frac{z}{1+z^2}+\frac{z^2}{1+z^4}+\frac{z^3}{1+z^6}=\frac{-2z^6}{z^6}=-2.$$故$$\left|\frac{z}{1+z^2}+\frac{z^2}{1+z^4}+\frac{z^3}{1+z^6}\right|=2.$$