Three, (50 points) Find all functions $f: \mathbf{R}_{+} \rightarrow \mathbf{R}_{+}$ such that for all positive numbers $x, y$, the following always holds:
$$
x y(f(x)+f(y)) \doteq(x+y) f(f(x) y) .
$$
Three, substituting $x=y=t$ into equation (1) yields
$$
2 t^{2} f(t)=2 t f(f(t) t) \text {, }
$$

which simplifies to $t f(t)=f(t f(t))\left(\forall t \in \mathbf{R}_{+}\right)$.
If there exist $x_{1} \neq x_{2}$ such that $f\left(x_{1}\right)=f\left(x_{2}\right)$, then substituting $x=x_{1}, y=x_{2}$ into equation (1) gives
$$
x_{1} x_{2}\left(f\left(x_{1}\right)+f\left(x_{2}\right)\right)=\left(x_{1}+x_{2}\right) f\left(f\left(x_{1}\right) x_{2}\right) \text {. }
$$
$$
\begin{array}{l}
\text { By } f\left(x_{1}\right)=f\left(x_{2}\right) \text { and equation (2), we have } \\
2 x_{1} x_{2} f\left(x_{2}\right)=\left(x_{1}+x_{2}\right) f\left(f\left(x_{2}\right) x_{2}\right),
\end{array}
$$

which simplifies to $2 x_{1} x_{2} f\left(x_{2}\right)=\left(x_{1}+x_{2}\right) x_{2} f\left(x_{2}\right)$.
Since $x_{2}>0, f\left(x_{2}\right)>0$, we have $2 x_{1}=x_{1}+x_{2}$, which implies $x_{1}=x_{2}$, contradicting $x_{1} \neq x_{2}$.
Therefore, for any $x_{1} \neq x_{2}$,
$$
f\left(x_{1}\right) \neq f\left(x_{2}\right) \text {. }
$$

Substituting $t=1$ into equation (2) gives $f(1)=f(f(1))$.
By equation (3), we have $f(1)=1$.
Substituting $x=1$ into equation (1) gives
$$
y(1+f(y))=(1+y) f(y) \text {. }
$$

Thus, $f(y)=y$.
Upon verification, $f(x)=x$ satisfies the problem's requirements.