6. Since $f\left(x_{1} x_{2}\right)=f\left(x_{1}\right)+f\left(x_{2}\right)$, let $x_{1}=x_{2}=1$, we get $f(1)=f(-1)=0$. Then let $x_{1}=x \cdot x_{2}=-1$, we get $f(x)=f(-x)$, so the function $f(x)$ is an even function.
Since $f(x)$ is increasing on $(0,+\infty)$, $f(x)$ is decreasing on $(-\infty, 0)$.
$f(x)+f\left(x-\frac{1}{2}\right)=f\left[x\left(x-\frac{1}{2}\right)\right] \leqslant 0, f\left[x\left(x-\frac{1}{2}\right)\right] \leqslant f(1)$ or $f\left[x\left(x-\frac{1}{2}\right)\right] \leqslant f(-1)$.
Therefore, $0<x\left(x-\frac{1}{2}\right) \leqslant 1$ or $-1<x\left(x-\frac{1}{2}\right) \leqslant 0$, thus the solution to the inequality is
$$
\left[\frac{1-\sqrt{17}}{4}, 0\right) \cup\left(0, \frac{1}{2}\right) \cup\left(\frac{1}{2}, \frac{1+\sqrt{17}}{4}\right] .
$$
6. (2006 National High School Mathematics Competition Henan Grade 1 Test) Let the function $f(x)(x \in \mathbf{R}, x \neq 0)$ for any non-zero real numbers $x_{1}, x_{2}$, have $f\left(x_{1} x_{2}\right)=f\left(x_{1}\right)+f\left(x_{2}\right)$, and $f(x)$ is an increasing function on $(0,+\infty)$, then the solution to the inequality $f(x)+$ $f\left(x-\frac{1}{2}\right) \leqslant 0$ is $\qquad$.