1. Determine all pairs of real numbers $(x, y)$, satisfying $x=3 x^{2} y-y^{3}, y=x^{3}-3 x y^{2}$.
1. Let $a=x+y \mathrm{i}, b=x-y \mathrm{i}$. Then
$$
\begin{array}{l}
a^{3}=(x+y \mathrm{i})^{3}=x^{3}+3 x^{2} y \mathrm{i}-3 x y^{2}-y^{3} \mathrm{i} \\
=y+x \mathrm{i}=\mathrm{i}(x-y \mathrm{i})=b \mathrm{i}, \\
b^{3}=(x-y \mathrm{i})^{3}=x^{3}-3 x^{2} y \mathrm{i}-3 x y^{2}+y^{3} \mathrm{i} \\
=y-x \mathrm{i}=-\mathrm{i}(x+y \mathrm{i})=-a \mathrm{i} .
\end{array}
$$

Thus $|b|=|a|^{3},|a|=|b|^{3}$
$\Rightarrow|a|=|b|=0$ or 1.
(1) If $|a|=|b|=0$, then $(x, y)=(0,0)$.
(2) If $|a|=|b|=1$, then
$$
\arg b+\frac{\pi}{2}-3 \arg a, \arg a+\frac{3 \pi}{2}-3 \arg b
$$

are integer multiples of $2 \pi$.
Since $a$ and $b$ are conjugate complex numbers, $\arg a+\arg b=2 \pi$.
Thus $\frac{\pi}{2}-4 \arg a$ is an integer multiple of $2 \pi$.
Therefore, $\arg a=\frac{\pi}{8}, \frac{5 \pi}{8}, \frac{9 \pi}{8}, \frac{13 \pi}{8}$,
$$
\begin{aligned}
(x, y)= & \left(\frac{\sqrt{2+\sqrt{2}}}{2}, \frac{\sqrt{2-\sqrt{2}}}{2}\right), \\
& \left(-\frac{\sqrt{2-\sqrt{2}}}{2}, \frac{\sqrt{2+\sqrt{2}}}{2}\right), \\
& \left(-\frac{\sqrt{2+\sqrt{2}}}{2},-\frac{\sqrt{2-\sqrt{2}}}{2}\right), \\
& \left(\frac{\sqrt{2-\sqrt{2}}}{2},-\frac{\sqrt{2+\sqrt{2}}}{2}\right) .
\end{aligned}
$$

In summary, $(x, y)=(0,0),\left(\frac{\sqrt{2+\sqrt{2}}}{2}, \frac{\sqrt{2-\sqrt{2}}}{2}\right)$,
$$
\begin{array}{l}
\left(-\frac{\sqrt{2-\sqrt{2}}}{2}, \frac{\sqrt{2+\sqrt{2}}}{2}\right), \\
\left(-\frac{\sqrt{2+\sqrt{2}}}{2},-\frac{\sqrt{2-\sqrt{2}}}{2}\right), \\
\left(\frac{\sqrt{2-\sqrt{2}}}{2},-\frac{\sqrt{2+\sqrt{2}}}{2}\right) .
\end{array}
$$