Áp dụng BĐT $\dfrac{1}{a+b+c}\le\dfrac{1}{9}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)$ ta có:$S=\dfrac{1}{\dfrac{1}{x}+\dfrac{2}{y}}+\dfrac{3}{\dfrac{1}{y}+\dfrac{2}{z}}+\dfrac{6}{\dfrac{1}{z}+\dfrac{2}{x}}$
$\le\dfrac{1}{9}(x+2y)+\dfrac{3}{9}(y+2z)+\dfrac{6}{9}(z+2x)$
$=\dfrac{1}{9}(13x+5y+12z)=1$
Vậy $\max S=1 \Leftrightarrow x=y=z=\dfrac{3}{10}$