$\begin{array}{l}\text { 9. } \frac{1-2 \log _{6} 5 \cdot \log _{10} 3 \cdot \log _{15} 2}{\log _{6} 5 \cdot \log _{10} 3+\log _{10} 3 \cdot \log _{15} 2+\log _{15} 2 \cdot \log _{6} 5} \\ =\end{array}$
9. 1.

Let $a=\log _{6} 5, b=\log _{10} 3, c=\log _{15} 2$. Then $a=\frac{\ln 5}{\ln 2+\ln 3}, b=\frac{\ln 3}{\ln 2+\ln 5}$,
$$
c=\frac{\ln 2}{\ln 3+\ln 5} .
$$

Thus, $\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}=1$.
Simplifying to $2 a b c+(a b+b c+c a)=1$.
Therefore, the original expression $=\frac{1-2 a b c}{a b+b c+c a}=1$.