Example 5 Find all real solutions $x, y$ of the system of equations
$$
\left\{\begin{array}{l}
x^{4}+y^{2}-x y^{3}-\frac{9}{8} x=0, \\
y^{4}+x^{2}-x^{3} y-\frac{9}{8} y=0
\end{array}\right.
$$

(1985, AIME Mathematics Competition)
Solving (1)-(2), factorizing, we get
$$
\begin{array}{l}
(x-y)\left(x^{3}+y^{3}+2 x^{2} y+2 x y^{2}-\frac{9}{8}\right) \\
=0 . \\
\therefore x-y=0
\end{array}
$$

or \(x^{3}+y^{3}+2 x^{2} y+2 x y^{2}-\frac{9}{8}=0\).
From (3), we get \(y=x\). Substituting into (1), and simplifying, we get \(x^{2}-\frac{9}{8} x=0\).
Solving, we get \(x_{1}=y_{1}=0, x_{2}=y_{2}=\frac{9}{8}\). \(x \cdot(4)-(1)\), factorizing, we get \(\left(x^{2}+x y+y^{2}\right)(2 x y-1)=0\). Since \(x^{2}+x y+y^{2}>0\), we have \(2 x y-1=0\) or \(y=\frac{1}{2 x}\).
Substituting (5) into (1), and simplifying, we get
$$
8 x^{6}-9 x^{3}+1=0 \text {. }
$$

Solving, we get \(x_{3}=1, x_{4}=\frac{1}{2}\). Substituting into (5), we get \(y_{3}=\frac{1}{2}, y_{4}=1\).
Therefore, the real solutions of the original system of equations are