<s>[INST] Functional Equation: Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $(x+y)(f(x)-f(y))=(x-y)f(x+y)$ Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$(x+y)(f(x)-f(y))=(x-y)f(x+y)$$
My attempt: 
If $x=-y \not = 0$ then $0= 2x f(0)$ so $f(0)=0$.

Suppose for the sake of contradiction that $f(x)=f(x+\epsilon)$ for some $x$ and $\epsilon>0$. Let $y=x+\epsilon$. Then $$0=\epsilon \cdot f(2x+\epsilon)$$ therefore as $2x+ \epsilon$ can take any real value $f$ is either strictly increasing or strictly decreasing or $f(x)=0 \; \; \forall \; \;x \in \mathbb{R}$. 
Note that $f(x)=ax$ is a solution $\forall \; \;a \in \mathbb{R}$. Thanks so much for any help!
 [/INST] $f(x)$ must be of the form $ax^2 + bx$.
Letting $x = 1$ and $y = 0$ in the equation, we find $f(0) = 0$.
Now define $g(x) = f(x)/x$ for $x \ne 0$. Taking $f(0) = 0$ for granted, the functional equation can be rewritten as 
$$(x-y)g(x + y) = xg(x) - yg(y), \qquad x, y, x+y \ne 0.$$
Substituting $1$ for $y$, we find
$$(x-1)g(x+1) = xg(x) - g(1), \qquad x \ne -1, 0.$$ 
Now substituting $x + 1$ for $x$ and $-1$ for $y$, we find
$$
\begin{align*}
(x+2)g(x) &= (x+1)g(x+1) + g(-1) &\quad \text{($x \ne -1,0$)}\\
(x-1)(x+2)g(x) &= (x+1)(x-1)g(x+1) + g(-1)(x-1) \\
(x^2 +x - 2)g(x) &= (x+1)[xg(x) - g(1)] + g(-1)(x-1) \\
-2g(x) &=-g(1)(x+1) +g(-1)(x-1).
\end{align*}
$$
The last relation is in fact true for all $x \ne 0$, including $x = -1$. 
This proves that the function $g(x)$ is linear, say $g(x) = ax + b$. Thus $f(x) = xg(x) = ax^2 + bx$. The relation $f(x) = ax^2 + bx$ is  valid for all $x$, including $x = 0$.
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