9. In a right-angled triangle $D E F$, the three vertices lie on the sides $A B, B C, C A$ of an equilateral triangle $A B C$, respectively, and $\angle D E F=$ $60^{\circ}, \angle E D F=30^{\circ}$,
Find the minimum value of $\frac{S_{\triangle D E F}}{S_{\triangle A B C}}$.
It is easy to prove that $\triangle B D E \sim \triangle C E F$, and the similarity ratio is 2. Let $\angle B D E = \alpha, D E = 2, B D = 2 x, B E = 2 y$, then the side length of $\triangle A B C$ is $2 y + x$,
Thus, $A D = 2 y - x$,
$A F = x + y$. Applying the Law of Sines in $\triangle A D F$, we have
$$
\begin{array}{l}
\frac{x + y}{\sin \left(150^{\circ} - \alpha\right)} = \frac{\sqrt{3}}{\sin 60^{\circ}} = 2, \\
\frac{2 y - x}{\sin \left(\alpha - 30^{\circ}\right)} = 2 \Rightarrow \left\{\begin{array}{l}
x + y = 2 \sin \left(150^{\circ} - \alpha\right), \\
2 y - x = 2 \sin \left(\alpha - 30^{\circ}\right).
\end{array}\right.
\end{array}
$$

Thus, $2 y + x = \frac{1}{3}[4(x + y) + (2 y - x)] = \frac{1}{3}\left[8 \sin \left(150^{\circ} - \alpha\right) + 2 \sin \left(\alpha - 30^{\circ}\right)\right]$ $= \frac{1}{3}(5 \sqrt{3} \sin \alpha + 3 \cos \alpha) \leqslant \frac{2 \sqrt{21}}{3}$, with equality when $\tan \alpha = \frac{5 \sqrt{3}}{3}$.
Therefore, the minimum value of $\frac{S_{\triangle D E F}}{S_{\triangle A B C}}$ is $\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{4}\left(\frac{2 \sqrt{21}}{3}\right)^{2}} = \frac{3}{14}$.