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!
 A: $f(x)$ 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 $x = 0$.
A: Note that $f(x) = a x^2 + b x$ is a solution.  I believe these are all the analytic solutions.
EDIT:
Yes, in fact they are all the differentiable solutions.
Taking $x=0$ we get $y f(0) = 0$, so $f(0) = 0$.  Now suppose $f$ is differentiable.  Taking the derivative of the equation with respect to $x$ and substituting $x=0$ we get
$$ - 2 f(y) + y f'(0) + y f'(y) = 0 $$
Letting $f'(0) = b$, the solutions of the differential equation $-2 f(y) + b y + y f'(y) = 0$ are  $f(y) = b y  + a y^2$ where $a$ is arbitrary.
EDIT:
Since $\dfrac{f(x) - f(y)}{x-y} = \dfrac{f(x+y)}{x+y}$, any solution that is continuous will be differentiable except possibly at $0$.  I am not at all convinced that every solution must be differentiable at $0$, or indeed that every solution must be continuous.
A: Hint:
First take $y=-x/2$ then
$$f(x) = 3f(x/2) + f(-x/2)$$
Taking $x\to -x$ we get
$$f(-x) = 3f(-x/2) + f(x/2)$$
This motivates us to define $g(x) = f(x) + f(-x)$ which by adding the equations above is found to satisfy
$$g(x) = 4g(x/2)$$
which is much easier to work with. We can for example by a simple inductive argument show that $g(2^n x) = 4^n g(x)$ for all $x$ and $n\in\mathbb{Z}$. A simple consequence of this is that if we know $g$ on any interval on the form $(2^{-n-1},2^{-n}]$ with $n\in\mathbb{Z}$ then we know it for all other values.
A: $f$ can be only of the form $ax^2+bx$.
To see this, we observe that for any $a$ and $b$, if we define $g(x):=f(x)-\big(ax^2+bx\big)$, then $g$ satisfies the same functional equation $(x+y)\big(g(x)-g(y)\big)=(x-y)g(x+y)$. Letting $a=\frac{f(1)+f(-1)}{2}$ and $b=\frac{f(1)-f(-1)}{2}$ we get $g(1)=g(-1)=0$. now putting $y=\pm1$ in the functional equation, we'll have:
$$(x+1)g(x)=(x-1)g(x+1)$$
$$(x-1)g(x)=(x+1)g(x-1)$$
Substituting $x+1$ for $x$ in the last equation, we'll have:
$$(x+2)g(x)=xg(x+1)$$
By using both equations above containing $g(x)$ and $g(x+1)$, we get:
$$x(x-1)g(x+1)=x(x+1)g(x)=(x-1)(x+2)g(x)$$
$$\therefore\big(x^2+x\big)g(x)=\big(x^2+x-2\big)g(x)$$
$$\therefore g(x)=0$$
Hence for every real number $x$, $f(x)$ is equal to $ax^2+bx$.
