, 2015

Define $g: \mathbb{R}_{>0} \rightarrow \mathbb{R}$ by $g(x)=1-f(x)$ for all $x>0$. The functional equation can be rewritten $g(x / y)=g(x) g(y)$ for all $x, y>0$.

Putting $y=x=1$, we get $g(1)=g(1)^2$. This means that $g(1)=0$ or $g(1)=1$.

1. Assume $g(1)=0$. For all $x>0$, we have $g(x)^2=g(x / x)=g(1)=0$. Hence $g(x)=0$ for all $x>0$. Conversely, $g=0$ is a solution of the given functional equation.

2. Assume $g(1)=1$. In this case, for all $x>0$, we have $g(x)=g(1) g(x)= g(1 / x)$.
Let $x>0$. We have
$$g(x)=g\left(\frac{\sqrt{x}}{1 / \sqrt{x}}\right)=g(\sqrt{x}) g(1 / \sqrt{x})=g(\sqrt{x})^2=g(\sqrt{x} / \sqrt{x})=g(1)=1 .$$
Conversely, $g=1$ is a solution of the given functional equation.

In conclusion, $f(x)$ is identically 0 or identically 1 .