Find all functions from $\mathbb{R}_{>0}$ to $\mathbb{R}_{>0}$ such that:

$$
f(x) f(y)=2 f(x+y f(x))
$$
If possible, we can try:

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

So if $f(x) \leq 1$, we can choose $x, y$ such that $y=x+y f(x)$, and thus $f(x)=2$, which is absurd! Therefore, $f(x)>1$ for all $x$. Let's take $x=y$. This gives:

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

Since $f(x+x f(x))$ is greater than $1$, $f(x)$ is greater than $\sqrt{2}$, so $f(x+x f(x))$ is greater than $\sqrt{2}$, and then $f(x)$ is greater than $\sqrt{2 \sqrt{2}}$. Continuing this, $f(x) \geq 2^{1 / 2+1 / 4+1 / 8+\ldots}=2$ for all $x$. As $y$ ranges over $\mathbb{R}$, $x+y f(x)$ ranges over the reals greater than $x$. Moreover, $f(y) \geq 2$, so for all $y$:

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

In other words, $f$ is increasing.

Suppose there exists a real $x_{0}$ such that $f(x_{0})=2$. In this case, with $x=y=x_{0}$, we have:

$$
f(x_{0})^{2}=2 f(x_{0}+x_{0} f(x_{0})) \Longrightarrow f(x_{0}+x_{0} f(x_{0}))=2
$$

We define the sequence $a_{n}$ by $a_{0}=x_{0}$ and $a_{n+1}=a_{n}+a_{n} f(a_{n})$. From the above, we have $f(a_{n})=2$ for all $n$, and since $f(a_{n}) \geq 2$, $(a_{n})$ diverges, which implies $f(a_{n})=2$ for all $n$, which implies that $f(x)=2$ for all $x$ by the increasing nature of $f$.

We note that $f(1)=2 \alpha$ with $\alpha>1$. We then observe that:

$$
\begin{gathered}
P(1,1) \Longleftrightarrow f(1)^{2}=2 f(1+f(1)) \Longrightarrow 2 f(1+2 \alpha)=4 \alpha^{2} \Longrightarrow f(1+2 \alpha)=2 \alpha^{2} \\
P(1+2 \alpha, 1) \Longleftrightarrow 2 \alpha \cdot 2 \alpha^{2}=2 f\left(1+2 \alpha+2 \alpha^{2}\right) \Longrightarrow f\left(1+2 \alpha+2 \alpha^{2}\right)=2 \alpha^{3} \\
P\left(1+2 \alpha+2 \alpha^{2}, 1\right) \Longleftrightarrow 2 \alpha \cdot 2 \alpha^{3}=2 f\left(1+2 \alpha+2 \alpha^{2}+2 \alpha^{3}\right) \Longrightarrow f\left(1+2 \alpha+2 \alpha^{2}+2 \alpha^{3}\right)=2 \alpha^{4}
\end{gathered}
$$

By immediate induction, we can deduce that $f\left(1+2 \alpha+2 \alpha^{2}+\ldots+2 \alpha^{n}\right)=2 \alpha^{n+1}$. In particular:

$$
\begin{gathered}
P\left(1+2 \alpha+2 \alpha^{2}+\ldots+2 \alpha^{n}, 1+2 \alpha+2 \alpha^{2}+\ldots+2 \alpha^{n}\right) \\
\Longleftrightarrow 4 \alpha^{n+2}=2 f\left(\left(1+2 \alpha+2 \alpha^{2}+\ldots+2 \alpha^{n}\right)\left(1+2 \alpha^{n+1}\right)\right) \\
\Longrightarrow 2 \alpha^{n+2}=f\left(4 \alpha^{2 n+1}+\ldots+1\right)
\end{gathered}
$$

For large $n$, $4 \alpha^{2 n+1}+\ldots+1>2 \alpha^{2 n+1}+2 \alpha^{2 n}+\ldots+1$, so:

$$
2 \alpha^{2 n+2}=f\left(4 \alpha^{2 n+1}+\ldots+1\right)>f\left(2 \alpha^{2 n+1}+2 \alpha^{2 n}+\ldots+1\right)=2 \alpha^{2 n+2}
$$

Absurd!

Key ideas: Use the bounds of the domain and improve them.