$6 \cdot 179$ Find all functions $f: R \rightarrow R$, such that
$$
x f(x)-y f(y)=(x-y) f(x+y) .
$$
[Solution] If $f$ is a real constant function, i.e., $f(x)$ is always equal to some real constant $b$, then it clearly satisfies the given functional equation.
Suppose $f$ is not a real constant function.
Let the line passing through the two points $(x, f(x))$ and $(y, f(y))$ be denoted as $l_{x, y}$. The equation of the line $l_{x, y}$ is
$$
\frac{y^{*}-f(x)}{x^{*}-x}=\frac{f(y)-f(x)}{y-x},
$$

where $\left(x^{*}, y^{*}\right)$ are the coordinates of a moving point on the line $l_{x, y}$. From the given condition, we have
$$
\begin{array}{l}
(x-y) f(x)+y f(x)-y f(y)=(x-y) f(x+y), \\
y(f(x)-f(y))=(x-y)(f(x+y)-f(x)),
\end{array}
$$

When $y \neq 0$, we have
$$
\frac{f(x+y)-f(x)}{(x+y)-x}=\frac{f(x)-f(y)}{x-y} .
$$

Therefore, the point $(x+y, f(x+y))$ lies on the line $l_{x, y}$.
Assume $x \neq 0$, then using the above result, we know that on the line $l_{x, 2 x}$, there are points $(3 x, f(x)), (4 x, f(x)), \cdots \cdots, (n x, f(n x)), n \in \mathbb{N}$. Similarly, on the line $l_{y, 2 y}$, there are points $(n y, f(n y)), n \in \mathbb{N}$.
Since the point $(x+y, f(x+y))$ lies on the line $l_{x, y}$, we have
$$
l_{x, y} \equiv l_{x, x+y} \equiv l_{y, x+y} .
$$

This implies that on the line $l_{x, y}$, there are points
$$
(2 x+y, f(2 x+y)), (2 y+x, f(2 y+x)) .
$$

Since $(2 x+y, f(2 x+y))$ lies on the line $l_{2 x, y}$, the point $(2 x, f(2 x))$ lies on
$l_{y, 2 x+y}$, which is the same as $l_{x, y}$. Therefore,
$$
l_{x, y} \equiv l_{x, 2 x} \text{. }
$$

Similarly,
$$
l_{x, y} \equiv l_{y, 2 y} .
$$

Let $l = l_{x, y}$. Since $x \neq 0$, taking $y = -x$, we have $x + y = 0$, so the line $l$ passes through the point $(0, f(0))$. Therefore, for any real number $x$, the point $(x, f(x))$ lies on this line $l$. Thus, there exist real numbers $a, b$ such that
$$
f(x) = a x + b, \quad a \neq 0 .
$$

It is easy to verify that when $f(x) = a x + b, a \neq 0$, the function satisfies the given functional equation.
In conclusion, all functions that satisfy the given functional equation are
$$
f(x) = a x + b, \quad a, b \in \mathbb{R} .
$$