9. Given three distinct real numbers $a, b, c$ form a geometric sequence, and $\log _{c} a, \log _{b} c, \log _{a} b$ form an arithmetic sequence with a common difference of $d$. Then $d=$ $\qquad$ .
9. $\frac{3}{2}$.

Since $a, b, c$ form a geometric sequence, we have $b^{2}=a c$.
Thus, $\log _{b} a+\log _{b} c=2$.
Let $\log _{b} a=x, \log _{b} c=y$. Then $x+y=2$.
Given that $\log _{c} a, \log _{b} c, \log _{a} b$ form an arithmetic sequence,
$\log _{c} a+\log _{a} b=2 \log _{b} c \Rightarrow \frac{x}{y}+\frac{1}{x}=2 y$.
Combining $x+y=2$ and $\frac{x}{y}+\frac{1}{x}=2 y$ and simplifying, we get
$2 x^{3}-9 x^{2}+9 x-2=0$
$\Rightarrow(x-1)\left(2 x^{2}-7 x+2\right)=0$
$\Rightarrow x=1$ or $\frac{7+\sqrt{33}}{4}$ or $\frac{7-\sqrt{33}}{4}$.
When $x=1$, $a=b$ (discard);
When $x=\frac{7+\sqrt{33}}{4}$, $y=\frac{1-\sqrt{33}}{4}$, in this case, the common difference is
$d=\log _{b} c-\log _{c} a=y-\frac{x}{y}=\frac{3}{2}$;
When $x=\frac{7-\sqrt{33}}{4}$, $y=\frac{1+\sqrt{33}}{4}$, in this case, the common difference is
$$
d=\log _{b} c-\log _{c} a=y-\frac{x}{y}=\frac{3}{2}.
$$

Therefore, the common difference $d=\frac{3}{2}$.