$9.244 \sqrt{1-9\left(\log _{1 / 8} x\right)^{2}}>1-4 \log _{1 / 8} x$.
Solution. $\sqrt{1-\log _{2}^{2} x}>1+\frac{4}{3} \log _{2} x$.

$$
\text { Let } y=\log _{2} x, \text { then } 3 \sqrt{1-y^{2}}>3+4 y
$$

1) $\left\{\begin{array}{l}3+4 y \geq 0, \\ 9\left(1-y^{2}\right)>(3+4 y)^{2}\end{array} \Leftrightarrow\left\{\begin{array}{l}y \geq-\frac{3}{4}, \\ y\left(y+\frac{24}{25}\right)<0\end{array} \Rightarrow y \in\left[-\frac{3}{4} ; 0\right)\right.\right.$.
2) $\left\{\begin{array}{l}3+4 y<0, \\ 1-y^{2} \geq 0\end{array} \Leftrightarrow y \in\left[-1 ;-\frac{3}{4}\right) . \quad\right.$ In the end: $\quad y \in[-1 ; 0) \Leftrightarrow-1 \leq \log _{2} x<0 \Rightarrow$ Answer: $\quad x \in\left[\frac{1}{2} ; 1\right)$.