See: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1249246&ml=1#p1249246

OK, I post link on this post too (in olympiad topic)

This solution assumes that $f(x)$ is a differentiable function over the reals. If we take $x = 1$ & $y = 0$, then we obtain: $(1 + 0)[f(1) - f(0)] = (1 - 0)f(1 + 0) \Rightarrow f(1) - f(0) = f(1) \Rightarrow f(0) = 0$. Differentiating our functional equation with respect to both $x$ and $y$ next yields: $f(x) - f(y) + (x+y)f'(x) = (x-y)f'(x+y) + f(x+y)$ (i) $f(x) - f(y) - (x+y)f'(y) = (x-y)f'(x+y) - f(x+y)$ (ii) If we now multiply (ii) by -1 and add it to (i), then we obtain: $[f'(x) + f'(y)](x + y) = 2f(x+y)$ (iii) and another differentiation with respect to $x$ and $y$ yields: $f''(x)(x+y) + f'(x) + f'(y) = 2f'(x+y)$ (iv) $f''(y)(x+y) + f'(x) + f'(y) = 2f'(x+y)$ (v) and equating (iv) and (v)) finally gives us: $f''(x) = f''(y) =$ constant $\Rightarrow f(x) = Ax^2 + Bx + C$ (vi). Since we've already determined $f(0) = 0 \Rightarrow C = 0$ and we now substitute $f(x) = Ax^2 + Bx$ back into the original functional equation: $(x+y)[Ax^2 + Bx - Ay^2 - By] = (x-y)[A(x+y)^2 + B(x+y)]$ (vii) and expanding both sides of (vii) gives: $Ax^3 + Bx^2 - Axy^2 - Bxy + Ax^2y + Bxy - Ay^3 - By^2 = Ax^3 + Axy^2 + 2Ax^2y + Bx^2 + Bxy - Ax^2y - Ay^3 - 2Axy^2 - Bxy - By^2$; or $Ax^3 + Bx^2 - Axy^2 - Bxy + Ax^2y + Bxy - Ay^3 - By^2 = Ax^3 + Ax^2y + Bx^2 + Bxy - Ay^3 - Axy^2 - Bxy - By^2$ which is equal on BOTH sides and is therefore independent of $A$ and $B$. Ultimately, the solution is the set of functions $f(x) = Ax^2 + Bx$ (for real constants $A$ & $B$).