6. Let $\mathbb{R}^{+}$ be the set of positive real numbers. Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that $f(x+f(y))=y f(x y+1)$, for all $x, y>0$.
Solution. Suppose there exists a real number $y$ such that $y-1$ and $f(y)-1$ are both non-zero and of the same sign. Then we can take the positive real number

$$
x=\frac{f(y)-1}{y-1}
$$

from which we get $f(y)=x(y-1)+1$ and $x+f(y)=x y+1$. Therefore, $f(x+f(y))=f(x y+1)$ and using the statement, we have

$$
y=\frac{f(x+f(y))}{f(x y+1)}=1
$$

This is a contradiction, and therefore, $y-1$ and $f(y)-1$ cannot both be non-zero and of the same sign. That is, $f(y) \geq 1$ for all $y \neq 1$.

For all $y>1$, let's take the real number $x=(y-1) / y>0$. Then, we have

$$
f\left(\frac{y-1}{y}+f(y)\right)=y f(y)
$$

Now we distinguish two cases:

- If $y f(y)>1$, then $\frac{y-1}{y}+f(y) \leq 1$, so $y f(y) \leq 1$, a contradiction.
- If $y f(y)=1$, then $f(y)=\frac{1}{y}$.

Now let $y \leq 1$, and take $x=1$. Then both $x+f(y)=1+f(y)$ and $x y+1=y+1$ are real numbers greater than 1, so

$$
\frac{1}{x+f(y)}=\frac{y}{x y+1} \Leftrightarrow x y+1=x y+y f(y)
$$

and we conclude that $f(y)=\frac{1}{y}$ also for all $y \leq 1$. Therefore, the only function that can satisfy the conditions of the statement is $f(y)=\frac{1}{y}$. We also see that

$$
f(x+f(y))=f\left(\frac{x y+1}{y}\right)=\frac{y}{x y+1}=y f(x y+1)
$$

so indeed $f(y)=\frac{1}{y}$ is a solution, and there can be no others.