9. find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that for all $x, y>0$ holds

$$
f(y f(x))(x+y)=x^{2}(f(x)+f(y))
$$

## 1st solution
Set $x=y$, then $f(x f(x)) follows \cdot 2 x=x^{2} \cdot 2 f(x)$. Because $x, f(x)>0$ this is equivalent to $f(x f(x))=x f(x)$ for all $x>0$. However, this means that $x f(x)$ is a fixed point of $f$ for all $x>0$. We now show that $f$ has only one fixed point and assume that $a$ and $b$ are two fixed points. Set $x=a$ and $y=b$, then because $f(a)=a$ and $f(b)=b$ the equation follows

$$
f(a b)(a+b)=a^{2}(a+b) \quad \Longleftrightarrow \quad f(a b)=a^{2}
$$

If you set $x=b$ and $y=a$, then analogously follows

$$
f(a b)(a+b)=b^{2}(a+b) \quad \Longleftrightarrow \quad f(a b)=b^{2}
$$

A comparison of (1) and (2) shows that $a=b$, therefore $f$ has only one fixed point $a$. In particular, $x f(x)=a$ for all $x>0$. The only possible solutions are of
the form $f(x)=a / x$. Substituting now gives $a=1$. The only solution to the equation is

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

## 2nd solution

We set $c=f(1)$. Inserting $x=y=1$ gives $f(c)=c$. Now set $x=c$ and $y=1$. With $f(c)=c$ it now follows that $f(c)(c+1)=c^{2}(c+c)$, i.e.

$$
c(c+1)=2 c^{3} \quad \Longleftrightarrow \quad c(c-1)\left(c+\frac{1}{2}\right)=0
$$

Because $c>0$, $f(1)=c=1$ must apply. Now set $x=1$, then follows

$$
f(y)(1+y)=1+f(y) \Longleftrightarrow f(y)=\frac{1}{y}, \quad \forall y>0
$$

Insert shows that this really is a solution.