9. Given that $a$ and $b$ are the roots of the equation
$$
\log _{3 x} 3+\log _{27} 3 x=-\frac{4}{3}
$$

then $a+b=$ $\qquad$
9. $\frac{10}{81}$.

The original equation is transformed into
$$
\begin{array}{l}
\frac{1}{1+\log _{3} x}+\frac{1+\log _{3} x}{3}=-\frac{4}{3} . \\
\text { Let } 1+\log _{3} x=t . \\
\text { Then } \frac{1}{t}+\frac{t}{3}=-\frac{4}{3} \Rightarrow t=-1 \text { or }-3 \\
\Rightarrow 1+\log _{3} x=-1 \text { or }-3 .
\end{array}
$$

Thus, the two roots of the equation are $\frac{1}{9}$ and $\frac{1}{81}$. Therefore, $a+b=\frac{10}{81}$.