Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x, y \in \mathbb{R}$,

$$
(x-y) f(x+y)=x f(x)-y f(y) .
$$
We claim that the answer is $f(t)=m t+b$ for $m, b \in \mathbb{R}$. This works because

$$
\begin{aligned}
m(x+y)(x-y)+b(x-y) & =m\left(x^{2}-y^{2}\right)+b x-b y \\
(x-y)(m(x+y)+b) & =x(m x+b)-y(m y+b) .
\end{aligned}
$$

We now show that only $m t+b$ works. Let $P(x, y)$ denote the problem equation. Taking $P(1, t-1)+P(t-1,-t)+P(-t, t)$ we get

$$
\begin{array}{rlrl}
P(1, t-1): & (2-t) f(t) & =f(1)-(t-1) f(t-1) \\
P(t-1,-t): & (2 t-1) f(-1) & =(t-1) f(t-1)+t f(-t) \\
+\quad P(-t, t): & -2 t f(0) & =-t f(-t)-t f(t) \\
\hline(2-t) f(t)+(2 t-1) f(-1)-2 t f(0) & =f(1)-t f(t) \\
2 f(t) & =f(1)+2 t f(0)+(1-2 t) f(-1) \\
f(t) & =(f(0)-f(-1)) t+\frac{f(1)+f(-1)}{2},
\end{array}
$$

so taking $m=f(0)-f(-1), b=\frac{f(1)+f(-1)}{2}$ gives $f(t)=m t+b$, as desired.