9. Given that $a, b$ are the roots of the equation $\log _{3 x} 3+\log _{27}(3 x)$ $=-\frac{4}{3}$, then $a+b=$ $\qquad$
$$
\begin{array}{l}
\frac{1}{1+\log _{3} x}+\frac{1}{3}\left(1+\log _{3} x\right)=-\frac{4}{3} \xlongequal{t=\log _{3} x} 3+(1+t)^{2}=-4(1+t) \Rightarrow t^{2}+6 t+8=0 \\
\Rightarrow t_{1}=-2, t_{2}=-4 \Rightarrow a=\frac{1}{9}, b=\frac{1}{81} \Rightarrow a+b=\frac{10}{81}
\end{array}
$$

The translation is as follows:

$$
\begin{array}{l}
\frac{1}{1+\log _{3} x}+\frac{1}{3}\left(1+\log _{3} x\right)=-\frac{4}{3} \xlongequal{t=\log _{3} x} 3+(1+t)^{2}=-4(1+t) \Rightarrow t^{2}+6 t+8=0 \\
\Rightarrow t_{1}=-2, t_{2}=-4 \Rightarrow a=\frac{1}{9}, b=\frac{1}{81} \Rightarrow a+b=\frac{10}{81}
\end{array}
$$