How to solve $f\left(\frac{f(x)}{yf(x)+1}\right)=\frac{x}{xf(y)+1}$? I'm currently working on the following functional equation:
Find all $f:\mathbb{R_{>0}}\to\mathbb{R_{>0}}$ such that for all $x,y\in\mathbb{R_{>0}}$:
$$
f\left(\frac{f(x)}{yf(x)+1}\right)=\frac{x}{xf(y)+1}
$$
I think, i have managed to show that $f$ is bijective, but how to proceed further?
 A: Denote $P(x,y)$ the assertion that the equation above holds for som $x,y$.
First, I will give a proof of bijectivity since you don't seem to feel safe with it.
First injectivity. Assume $f(a)=f(b)$ for some $a,b$. Then comparing $P(a,y)$ and $P(b,y)$ we obtain $\frac{a}{af(y)+1}=\frac{b}{bf(y)+1}$ and hence $abf(y)+b=abf(y)+a$ and hence $a=b$ i.e. $f$ is injective.
Now, assume $f$ assumes the value $k$ i.e. $f(x_0)=k$ for some $x_0$.
Then $P(x,x_0)$ as $x$ runs through all positive numbers shows that $f$ obtains all numbers between 0 and $\frac{1}{k}$.
So if we take any $k \in f(\mathbb{R})$ (satisfying $k>0$) we conclude that every positive number $\epsilon$ sufficiently close to 0 is assumed.
But then again applying the same trick we obtain that all numbers between $0$ and $\frac{1}{\epsilon}$ are obtained where $\frac{1}{\epsilon}$ gets arbitrarily large i.e. $f$ is surjective, thus bijective.
Now we are proving that $f(x)=x$ holds for all $x$.
First, we prove $f(x) \le x$ must hold.
Assume on the contrary that $f(a)=b>a$ for some $a,b>0$.
Then $P(a,\frac{1}{a}-\frac{1}{b})$ implies $b=\frac{a}{af(\frac{1}{a}-\frac{1}{b})+1}<\frac{a}{1}=a$ and hence $b<a$ contradicting the assumption.
So $f(x) \le x$ holds for all $x$.
Assume $f(a)=b<a$ holds for some $a,b>0$.
Surjectivity implies that there is some $y_0$ such that $f(y_0)=\frac{1}{a}-\frac{1}{b}$.
Then $P(a,y_0)$ implies $f(\frac{b}{y_0b+1})=b=f(a)$ and hence by injectivity we obtain $a=\frac{b}{y_0b+1}<\frac{b}{1}=b$ and hence $a<b$ contradicting the assumption.
Thus, we have proved that $f(x)=x$ must hold for all $x$ and this is indeed a solution.
