Czech-Polish-Slovak Mathematical Match

Suppose $f(t)=0$ for some $t>0$. For $0<x<t$ we choose $y=t-x>0$ and find $f(x)=0$. From setting $x=y=1$ we conclude that $f(1)\ne -1$. Hence by setting $x=y$ we get $f(x)-f(2x)=f(1)f(2x)$ for $x>0$. Inductively we find $$f(2^{n}x)=f(x)(1+f(1){)}^{-n}.(2)$$ Hence for an arbitrary $x>0$ we choose $n\in \mathbb{N}$ such that $\frac{x}{2^n}\le t$ and we conclude from (2) that $f(x)=0$.

Now suppose $f(t)\ne 0$ for all $t>0$. We define $g(x):=f(x{)}^{-1}$, $x>0$, and rewrite the given equation as $$g(x+y)-g(x)=\frac{g(x)}{g\left(\frac{x}{y}\right)}\text{for all }x,y>0.(3)$$ By setting $y=1$ we obtain $g(x+1)=g(x)+1$ for all $x>0$. It follows that $g(n)=n-1+g(1)$ holds for all $n\in \mathbb{N}$. Setting $x=y=2$ in (3) we now find $g(1)=1$. Setting $x=1$ we obtain $g(y)g(\frac{1}{y})=1$ for all $y>0$. We can now rewrite (3) as $$g(x+y)=g(x)+g(x)g\left(\frac{y}{x}\right)=g(x)\left(1+g\left(\frac{y}{x}\right)\right)=g(x)g\left(\frac{x+y}{x}\right).(4)$$ From (4) we directly see that $g(a)g(b)=g(ab)$ for all $a>0$ and $b>1$. Using $g(y)g(\frac{1}{y})=1$, we can rewrite (4) as $$g\left(\frac{x}{x+y}\right)g(x+y)=g(x),$$ which together with the previous line shows that $g(a)g(b)=g(ab)$ holds indeed for all $a,b>0$. In particular we see that $g(a)=g(\sqrt{a}{)}^2>0$, so $g$ is a positive function. Also, from the first equation in (4) we now infer the functional equation $$g(x+y)=g(x)+g(y)\text{for all }x,y\in (0,\infty ).$$ It is well known that this implies $g(x)=xg(1)=x$ for $x\in {\mathbb{Q}}_{>0}$. Since $g$ is positive, by (3) we deduce that $g(x+y)>g(x)$, so $g$ is strictly increasing. Hence $g(x)=x$ for all $x>0$.

Since $f(x)=0$ and $f(x)=\frac{1}{x}$ obviously satisfy the given equation, we have found all solutions. $\square$