7. Given the function $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for any real numbers $x, y$ it satisfies
$$
f(f(x)+f(y))=(x+y) f(x+y) .
$$

Find $f(x)$.
7. Taking $x=y=0$, we get $f(2 f(0))=0$.

Taking $x=2 f(0), y=0$, we get $f(f(0))=0$.
Taking $x=y=f(0)$, we get $f(0)=0$.
For any $x$, taking $y=0$, we get
$$
f(f(x))=x f(x) \text {. }
$$

Assume there exist $a, b \in \mathbf{R}$, satisfying
$$
f(a)=f(b) \neq 0 \text {. }
$$

Then $a f(a)=f(f(a))=f(f(b))=b f(b)$
$$
\Rightarrow a=b \text {. }
$$

Thus, if $f(x)$ can take the same value, it can only be 0.
Next, we prove: there exists $c \neq 0$ such that $f(c)=0$.
In equation (1), taking $y=1-x$, we get
$$
f(f(x)+f(1-x))=f(1) \text {. }
$$

If $f(1)=0$, then $c=1$, otherwise,
$$
f(x)+f(1-x)=1
$$

holds for any real number $x$.
Taking $x=0, x=\frac{1}{2}$, we get
$$
f(1)=1, f\left(\frac{1}{2}\right)=\frac{1}{2} \text {. }
$$

Then in equation (1), taking $x=1, y=\frac{1}{2}$, we get
$$
f\left(\frac{3}{2}\right)=\frac{3}{2} f\left(\frac{2}{3}\right) \Rightarrow f\left(\frac{3}{2}\right)=0 \text {. }
$$

Let $c$ be any non-zero real number such that $f(c)=0$.
In equation (1), taking $x=y=c$, we get $f(2 c)=0$.
Assume there exists a real number $t$ such that $f(t) \neq 0$, then in equation (1), taking $x=c, y=-c+t$ and taking $x=2 c, y=-2 c+t$, we get
$$
\begin{array}{l}
f(f(-c+t))=t f(t) \\
\quad=f(f(-2 c+t))(t f(t) \neq 0) . \\
\text { Hence } f(-c+t)=f(-2 c+t) \neq 0 \\
\Rightarrow-c+t=-2 c+t,
\end{array}
$$

which contradicts $c \neq 0$.
Therefore, such $t$ does not exist, i.e., for any real number $x$, we have $f(x)=0$.