$6 \cdot 179$ 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.

Assume $f$ 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 $l_{x, 2x}$, there are points $(3x, f(x)), (4x, f(x)), \cdots \cdots, (nx, f(nx)), n \in \mathbb{N}$. Similarly, on the line $l_{y, 2y}$, there are points $(ny, f(ny)), n \in \mathbb{N}$.

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
$$(2x+y, f(2x+y)), (2y+x, f(2y+x))$$

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

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

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
$$f(x) = ax + b, \quad a \neq 0$$

It is easy to verify that when $f(x) = ax + b, a \neq 0$, it satisfies the given functional equation.

In conclusion, all functions that satisfy the given functional equation are
$$f(x) = ax + b, \quad a, b \in \mathbb{R}$$