11. (20 points) Given non-zero complex numbers $x, y$ satisfy
$$
\begin{array}{l}
y^{2}\left(x^{2}-x y+y^{2}\right)+x^{3}(x-y)=0 . \\
\text { Find } \sum_{m=0}^{29} \sum_{n=0}^{29} x^{18 m n} y^{-18 m n} \text { . }
\end{array}
$$
11. Divide both sides of the known equation by $y^{4}$, $\left(\frac{x}{y}\right)^{4}-\left(\frac{x}{y}\right)^{3}+\left(\frac{x}{y}\right)^{2}-\left(\frac{x}{y}\right)+1=0$.
Let $\frac{x}{y}=\omega$, then
$$
\begin{array}{l}
\omega^{4}-\omega^{3}+\omega^{2}-\omega+1=0 \\
\Rightarrow \omega^{5}=-1 \Rightarrow \omega^{10}=\left(\omega^{2}\right)^{5}=1 .
\end{array}
$$

Thus, $\omega^{2}$ is a fifth root of unity.
Let $z=x^{18} y^{-18}=\left(\left(\frac{x}{y}\right)^{2}\right)^{9}=\left(\omega^{2}\right)^{9}$, which is still a fifth root of unity.
$$
\text { Also, } \sum_{m=0}^{29} \sum_{n=0}^{29} x^{18 m n} y^{-18 m n}=\sum_{m=0}^{29} \sum_{n=0}^{29} z^{m n} \text {, }
$$

When $5 \mid m$, $z^{m n}=1$;
When $5 \nmid m$, $z^{m}$ is still a fifth root of unity, and
$$
\sum_{k=0}^{4} z^{k m}=0,
$$

Therefore, $\sum_{m=0}^{29} \sum_{n=0}^{29} z^{m n}=\sum_{5 \mid m}^{29} \sum_{n=0}^{29} z^{m n}+\sum_{5 \nmid m}^{29} \sum_{n=0}^{29} z^{m n}$
$$
=6 \times 30+0=180 \text {. }
$$