The function $f(x)$ satisfies for any positive numbers $x, y$: $f(x y)=f(x)+f(y)$, and $f(2)=1$, then the value of $f\left(\frac{1}{64}\right)$ is $\qquad$ .
9. -6 .

From $f(1 \times 1)=f(1)+f(1)$, we get $f(1)=0$. $f(4)=f(2)+f(2)=2$, $f(8)=f(2)+f(4)=3$, $f(64)=f(8)+f(8)=6$. Since $0=f(1)=f\left(64 \times-\frac{1}{64}\right)=f(64)+f\left(\frac{1}{64}\right)$. Therefore, $f\left(\frac{1}{64}\right)=-6$.