9. As shown in Figure 4, in $\triangle A B C$, points $E$ and $F$ are on sides $A B$ and $A C$ respectively, and $E F / / B C$. If $S_{\triangle A B C}=1, S_{\triangle A E F}=$ $2 S_{\triangle E B C}$, then $S_{\triangle C E F}=$ $\qquad$
$9.3 \sqrt{3}-5$.
Let $\frac{A F}{F C}=t$. Then $S_{\triangle S B F}=\frac{C F}{F A} S_{\triangle A E F}=\frac{1}{t} S_{\triangle \triangle S F F}$.
Since $\triangle A E F \backsim \triangle A B C$, we have
$S_{\triangle \triangle E F}=\left(\frac{A F}{A C}\right)^{2} S_{\triangle \triangle B C}=\left(\frac{t}{1+t}\right)^{2}$.
Also, $S_{\triangle S B C}=\frac{B E}{B A} S_{\triangle \triangle B C}=\frac{1}{1+t}$, from the problem we get $\left(\frac{t}{1+t}\right)^{2}=\frac{2}{1+t}, t=1+\sqrt{3}$.
Therefore, $S_{\triangle C B F}=\frac{1}{t} S_{\triangle A E P}=\frac{1}{t} \times\left(\frac{t}{1+t}\right)^{2}$ $=\frac{t}{(1+t)^{2}}=3 \sqrt{3}-5$.