9. Let the function $f(x)=\frac{a x}{2 x+3}$. If $f(f(x))=x$ always holds, then the value of the real number $a$ is $\qquad$ .
II. 9. -3.
According to the problem, we know that $\frac{a \cdot \frac{a x}{2 x+3}}{2 \times \frac{a x}{2 x+3}+3}=x$, which means
$$
\begin{array}{c}
2 \times \frac{a x}{2 x+3}+3 \\
(2 a+6) x^{2}+9 x=a^{2} x . \\
\text { Therefore }\left\{\begin{array}{l}
2 a+6=0, \\
9=a^{2}
\end{array} \Rightarrow a=-3 .\right.
\end{array}
$$