2. Let $\mathbf{R}_{+}$ denote the set of positive real numbers. Find all functions $f: \mathbf{R}_{+} \rightarrow \mathbf{R}_{+}$ such that for all positive real numbers $x, y$, we have $f(x) f(y)=2 f(x+y f(x))$.
2. $f(x)=2$.

First, we prove: the function $f(x)$ is non-decreasing.
If there exist positive real numbers $x>z$, such that $f(x)0$, then $x+y f(x)=z+y f(z)$. Thus, we have
$$
\begin{array}{l}
f(x) f(y)=2 f(x+y f(x)) \\
=2 f(z+y f(z))=f(z) f(y),
\end{array}
$$

which implies $f(x)=f(z)$, a contradiction.
If $f(x)$ is not strictly increasing, then there exist positive real numbers $x>z$, such that $f(x)=f(z)$.
If $y \in\left(0, \frac{x-z}{f(x)}\right]$, then $z2 f(x) .
$$

Thus, $f(y)>2$.
Furthermore, for all $x>0$, we have
$$
\begin{array}{l}
2 f(x+1 \times f(x))=f(x) f(1) \\
=f(1) f(x)=2 f(1+x f(1)) .
\end{array}
$$

By the strict monotonicity of $f(x)$, we know
$$
x+1 \times f(x)=1+x f(1),
$$

which implies $f(x)=x(f(1)-1)+1$.
By choosing a sufficiently small $x$, we can make $f(x)0$ such that $f(y)>2$ leads to a contradiction.
In conclusion, the function that satisfies the conditions is $f(x)=2$.