Let $\mathbf{R}$ be the set of all real numbers. Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ such that for any $x, y \in \mathbf{R}$, we have
$$
f(x f(y))=x y f(x+y) .
$$
Given the problem, for any $x, y \in \mathbf{R}$, we have
$$
f(x f(y)) = x y f(x + y).
$$

Taking $x = 0$ in equation (1), we get $f(0) = 0$.
Taking $x = -y$ in equation (1), we get
$$
f(-y f(y)) = 0.
$$

Assume there exists an $x_{0} \in \mathbf{R}$, and $x_{0} \neq 0$, such that $f(x_{0}) \neq 0$.
Taking $y$ as $-y f(y)$ in equation (1), and using $f(0) = 0$ and equation (2), we get
$$
0 = x(-y f(y)) f(x - y f(y)).
$$

Substituting $y = x_{0}$ in equation (3), we get
$$
0 = x(-x_{0} f(x_{0})) f(x - x_{0} f(x_{0})).
$$

Substituting $x = x_{0} f(x_{0}) + x_{0}$ in equation (4), we get
$$
0 = -x_{0} f(x_{0}) x_{0}(f(x_{0}) + 1) f(x_{0}).
$$

Since $x_{0} \neq 0$ and $f(x_{0}) \neq 0$, it must be that $f(x_{0}) + 1 = 0$, i.e.,
$$
f(x_{0}) = -1.
$$

Substituting $y = x_{0}$ in equation (2) and using equation (5), we get $f(x_{0}) = 0$, which contradicts the assumption that $f(x_{0}) \neq 0$.
Therefore, for every $x \in \mathbf{R}$, we have $f(x) = 0$.