11.32. Solve the system of equations

$$
\left\{\begin{array}{l}
2 x+x^{2} y=y \\
2 y+y^{2} z=z \\
2 z+z^{2} x=x
\end{array}\right.
$$
11.32. The first equation can be rewritten as $y=\frac{2 x}{1-x^{2}}$ (it is clear that $x \neq \pm 1$). Let $x=\operatorname{tg} \varphi$. Then $y=\operatorname{tg} 2 \varphi$. Similarly, $z=\operatorname{tg} 4 \varphi$ and $x=\operatorname{tg} 8 \varphi$. Thus, $\operatorname{tg} 8 \varphi=\operatorname{tg} \varphi$. Therefore,

$$
\operatorname{tg} 7 \varphi=\frac{\operatorname{tg} 8 \varphi-\operatorname{tg} \varphi}{1+\operatorname{tg} 8 \varphi \operatorname{tg} \varphi}=\frac{\operatorname{tg} \varphi-\operatorname{tg} \varphi}{1+\operatorname{tg}^{2} \varphi}=0
$$

i.e., $7 \varphi=k \pi$, where $k$ is an integer. Conversely, if $7 \varphi=k \pi$, then $\operatorname{tg} 8 \varphi=\operatorname{tg} \varphi$ and we obtain a solution to the given system of equations. In total, we get 7 different solutions corresponding to the angles $\varphi=0, \frac{\pi}{7}, \ldots, \frac{6 \pi}{7}$.