1. Find all real pairs $(x, y)$ that satisfy:
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+x+y=x y(x+y)-\frac{10}{27}, \\
|x y| \leqslant \frac{25}{9} .
\end{array}\right.
$$
1. $(x, y)=\left(-\frac{1}{3},-\frac{1}{3}\right),\left(\frac{5}{3}, \frac{5}{3}\right)$.

From $x^{2}+y^{2}+x+y-x y(x+y)-2$
$$
\begin{array}{l}
=-(1-x)(1-y)(x+y+2) \\
\Rightarrow(1-x)(1-y)(x+y+2)=\frac{64}{27} .
\end{array}
$$

Let $k^{3}=1-x, l^{3}=1-y, m^{3}=x+y+2$. Then
$k^{3}+l^{3}+m^{3}=4, k l m=\frac{4}{3}$
$\Rightarrow k^{3}+l^{3}+m^{3}=3 k l m$
$\Rightarrow(k+l+m)\left((k-l)^{2}+(l-m)^{2}+(m-k)^{2}\right)=0$
$\Rightarrow k+l+m=0$ or $k=l=m$.
If $k=l=m$, then $x=y=-\frac{1}{3}$.
If $k+l+m=0$, by $k l m=\frac{4}{3}$, we get
$$
k l(k+l)=-\frac{4}{3} .
$$

Consider the condition
$$
|x y|=\left|\left(1-k^{3}\right)\left(1-l^{3}\right)\right| \leqslant \frac{25}{9} \text {. }
$$

Let $u=k+l, v=k l$. Then
$$
u v=-\frac{4}{3} \text {, }
$$
$\left|\left(1-k^{3}\right)\left(1-l^{3}\right)\right|=\left|u^{3}-v^{3}+3\right| \leqslant \frac{25}{9}$.
Thus, $-\frac{52}{9} \leqslant u^{3}-v^{3}=u^{3}+\frac{64}{27 u^{3}} \leqslant-\frac{2}{9}$.
By $(k+l)^{2} \geqslant 4 k l$
$\Rightarrow u^{2} \geqslant 4 v=-\frac{16}{3 u}$.
From equation (1) we know $u0 \\
\Rightarrow u^{3}+\frac{64}{27 u^{3}}+\frac{52}{9}=\left(u^{3}+\frac{16}{3}\right)\left(\frac{4}{9 u^{3}}+1\right) \leqslant 0 .
\end{array}
$$

Thus, $u^{3} \leqslant-\frac{16}{3}$

From equation (1) we know the equality holds, and
$$
u^{3}=-\frac{16}{3}, u^{2}=4 v \text {, }
$$

This indicates that $k=l=-\sqrt[3]{\frac{2}{3}}, x=y=\frac{5}{3}$.