Ta có: \(xy\cdot yz\cdot zx=\dfrac{2}{3}\cdot\dfrac{3}{5}\cdot\dfrac{2}{5}=\dfrac{4}{25}\)
\(\Leftrightarrow\left(xyz\right)^2=\dfrac{4}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}xyz=\dfrac{2}{5}\\xyz=-\dfrac{2}{5}\end{matrix}\right.\)
Trường hợp 1: \(xyz=\dfrac{2}{5}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3}\cdot z=\dfrac{2}{5}\\x\cdot\dfrac{3}{5}=\dfrac{2}{5}\\y\cdot\dfrac{2}{5}=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}z=\dfrac{3}{5}\\x=\dfrac{2}{3}\\y=1\end{matrix}\right.\)
Trường hợp 2: \(xyz=-\dfrac{2}{5}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3}\cdot z=-\dfrac{2}{5}\\x\cdot\dfrac{3}{5}=-\dfrac{2}{5}\\y\cdot\dfrac{2}{5}=-\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}z=-\dfrac{3}{5}\\x=-\dfrac{2}{3}\\y=-1\end{matrix}\right.\)
Vậy: \(\left(x,y,z\right)\in\left\{\left(\dfrac{2}{3};1;\dfrac{3}{5}\right);\left(-\dfrac{2}{3};-1;-\dfrac{3}{5}\right)\right\}\)