9.063. $4^{\frac{1}{x}-1}-2^{\frac{1}{x}-2}-3 \leq 0$.
Solution.

Domain of definition: $x \neq 0$.

From the condition $\frac{\left(2^{\frac{1}{x}}\right)^{2}}{4}-\frac{2^{\frac{1}{x}}}{4}-3 \leq 0 \Leftrightarrow\left(2^{\frac{1}{x}}\right)^{2}-2^{\frac{1}{x}}-12 \leq 0$, solve it as a quadratic equation in terms of $2^{\frac{1}{x}}$. We get $2^{\frac{1}{x}} \geq-3$, from which we find
$x \neq 0$, or $2^{\frac{1}{x}} \leq 4$, from which we find $\frac{1}{x} \leq 2, \frac{1}{x}-2 \leq 0, \frac{1-2 x}{x} \leq 0$, $\left\{\begin{array}{l}(2 x-1) x \geq 0, \\ x \neq 0,\end{array}\left[\begin{array}{l}x \geq \frac{1}{2}, \\ x<0 .\end{array}\right.\right.$

Answer: $\quad x \in(-\infty ; 0) \cup\left[\frac{1}{2} ; \infty\right)$