Example 12. Solve the system of equations
$$
\left\{\begin{array}{l}
x z=3 u, \\
u^{2}+y^{2}+y u=x^{2}, \\
y^{2}+z^{2}-y z=9, \\
x^{2}=u^{2}+z^{2}+2 u z+9
\end{array}\right.
$$

for positive solutions.
The original system of equations can be transformed into
$$
\left\{\begin{array}{l}
\frac{x}{y}=\frac{3}{u}, \\
u^{2}+y^{2}-2 u y \cos 120^{\circ}=x^{2}, \\
y^{2}+z^{2}-2 y z \cos 60^{\circ}=3^{2}, \\
x^{2}=(z+u)^{2}+3^{2} .
\end{array}\right.
$$

According to this, we can construct $\triangle A B C$, such that $\angle B=90^{\circ}$, and $A D$ bisects $\angle A$, as shown in the figure. Thus, $\angle B A D=\angle D A C=\angle C=30^{\circ}$. Therefore,
$$
\begin{array}{c}
z=B D=A B \operatorname{tg} 30^{\circ} \\
=3 \times \frac{\sqrt{3}}{3}=\sqrt{3} \\
y=\sqrt{3^{2}+(\sqrt{3})^{2}}=2 \sqrt{3} \\
u=y=2 \sqrt{3}, x=2 A B=2 \times 3=6
\end{array}
$$