jmc

Note that $$f(600)=f\left(500\cdot \frac{6}{5}\right)=\frac{f(500)}{6/5}=\frac{3}{6/5}=\boxed{\frac{5}{2}}.$$$$\text{OR}$$For all positive $x$, $$f(x)=f(1\cdot x)=\frac{f(1)}{x},$$so $xf(x)$ is the constant $f(1)$. Therefore, $$600f(600)=500f(500)=500(3)=1500,$$so $f(600)=\frac{1500}{600}=\boxed{\frac{5}{2}}$.

Note: $f(x)=\frac{1500}{x}$ is the unique function satisfying the given conditions.