9. In the acute $\triangle A B C$,
$$
\sin (A+B)=\frac{3}{5}, \sin (A-B)=\frac{1}{5}, A B=3 \text {. }
$$

Then the area of $\triangle A B C$ is $\qquad$
9. 9. $\frac{3(\sqrt{6}+2)}{2}$.

From the given information,
$$
\begin{array}{l}
\left\{\begin{array}{l}
\sin A \cdot \cos B+\cos A \cdot \sin B=\frac{3}{5}, \\
\sin A \cdot \cos B-\cos A \cdot \sin B=\frac{1}{5}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
\sin A \cdot \cos B=\frac{2}{5}, \\
\cos A \cdot \sin B=\frac{1}{5}
\end{array}\right. \\
\Rightarrow \tan A=2 \tan B .
\end{array}
$$

Also, $\tan (A+B)=\frac{\sin (A+B)}{\cos (A+B)}=-\frac{3}{4}$, and
$$
\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}=\frac{3 \tan B}{1-2 \tan ^{2} B} \text {, }
$$

Thus, $\tan B=\frac{\sqrt{6}+2}{2}$.
Let the height from vertex $A$ to side $B$ be $h$. Then
$$
\begin{array}{l}
3=A B=\frac{h}{\tan A}+\frac{h}{\tan B}=\frac{3 h}{2 \tan B} \\
\Rightarrow h=2 \tan B . \\
\text { Therefore, } S_{\triangle A B C}=\frac{1}{2} A B \cdot h=\frac{1}{2} \times 3 \times 2 \tan B \\
=3 \tan B=\frac{3(2+\sqrt{6})}{2} .
\end{array}
$$