Cesko-Slovacko-Poljsko

Simple manipulation with the given equation leads to $$\begin{array}{rl}1+yf(x)-yf(x+y)-y^{2}f(x)f(x+y)& =1,\\ yf(x)-yf(x+y)& =y^{2}f(x)f(x+y).\end{array}$$ After cancelling $y\ne 0$ another manipulation gives $$\begin{array}{rl}f(x)-f(x+y)& =yf(x)f(x+y),\\ \frac{f(x)-f(x+y)}{f(x+y)}& =yf(x),\\ \frac{f(x)}{f(x+y)}-1& =yf(x),\\ \frac{f(x)}{f(x+y)}& =1+yf(x),\\ f(x+y)& =\frac{f(x)}{1+yf(x)},\\ \frac{1}{f(x+y)}& =(1+yf(x))\cdot \frac{1}{f(x)}=\frac{1}{f(x)}+y.\end{array}$$ (evidently, all the cancelled expressions are nonzero) and so $$\frac{1}{f(x+y)}=y+\frac{1}{f(x)}$$ for any $x,y\in {\mathbb{R}}^{+}$. Hence $$y+\frac{1}{f(x)}=x+\frac{1}{f(y)}.$$ Setting $y=1$ we get $$\frac{1}{f(x)}=x+\frac{1}{f(1)}-1=x+c,\text{so}f(x)=\frac{1}{x+c}$$ with a constant $c$. As $f(x)>0$ for any $x>0$, we have $x+c>0$ for any $x>0$ and therefore $c\ge 0$. We can easily check that the function $f(x)=1/(x+c)$ satisfies the given conditions for any $c\ge 0$: $$\left(1+\frac{y}{x+c}\right)\left(1-\frac{y}{x+y+c}\right)=\frac{x+c+y}{x+c}\cdot \frac{x+y+c-y}{x+y+c}=\frac{x+c+y}{x+c}\cdot \frac{x+c}{x+y+c}=\frac{x+c+y}{x+y+c}.$$ But $\frac{x+c+y}{x+y+c}=1$, so the condition is satisfied for all $x,y>0$ and $c\ge 0$.

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Alternative solution.

Put $x=1$ and $f(1)=a>0$. Then we have $$(1+ay)(1-yf(y+1))=1,$$ $$ay-yf(y+1)(1+ay)=0,$$ $$f(y+1)=\frac{a}{1+ay}$$ Now put $y=1$ and $f(x+1)=a/(1+ax)$ to get $$(1+f(x))\left(1-\frac{a}{1+ax}\right)=1,$$ $$f(x)\cdot \frac{1+ax-a}{1+ax}=\frac{a}{1+ax},$$ $$f(x)=\frac{a}{1+ax-a}=\frac{1}{x+\frac{1}{a}-1}=\frac{1}{x+c}.$$ As in the first solution, we have $c\ge 0$ and it is easy to check that such a function satisfies the required conditions.