1. Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for all real numbers $x, y$, we have
$$
\begin{array}{l}
f(2 x y)+f(f(x+y)) \\
=x f(y)+y f(x)+f(x+y) .
\end{array}
$$
1. In equation (1), let $y=\frac{1}{2}$, we get
$$
\begin{array}{l}
f(x)+f\left(f\left(x+\frac{1}{2}\right)\right) \\
=x f\left(\frac{1}{2}\right)+\frac{1}{2} f(x)+f\left(x+\frac{1}{2}\right) .
\end{array}
$$

In equation (1), let $y=0$, and replace $x$ with $x+\frac{1}{2}$, we get
$$
\begin{aligned}
f(0) & +f\left(f\left(x+\frac{1}{2}\right)\right) \\
& =\left(x+\frac{1}{2}\right) f(0)+f\left(x+\frac{1}{2}\right) .
\end{aligned}
$$

Subtract the above two equations and rearrange to get
$$
\begin{array}{l}
f(x)-f(0)=2 x f\left(\frac{1}{2}\right)-2 x f(0) \\
\Rightarrow f(x)=2 x\left(f\left(\frac{1}{2}\right)-f(0)\right)+f(0) .
\end{array}
$$

Therefore, $f(x)$ must be a polynomial of degree no more than 1. Let $f(x)=a x+b$, substituting into equation (1) yields the complete solution to the original problem:
$$
f(x)=0 \text { or } f(x)=x \text { or } f(x)=2-x \text {. }
$$