The functional equation $f\bigl(x+yf(x)\bigr)+f\bigl(xf(y)-y\bigr) = f(x)-f(y)+2xy^2$

Question

Let $f:\mathbb{R} \to \mathbb{R}$ such that: $$f\bigl(x+yf(x)\bigr)+f\bigl(xf(y)-y\bigr) = f(x)-f(y)+2xy^2$$

I don't know how to solve this functional equation. Yet here is what I've noticed so far:

1. $f(0) = 0$ after the substitution: $(x,y) = (0,0)$
2. $f(-x) = -f(x)$ after the substitution: $(x,y) = (0,x)$
3. $f\bigl(x+xf(x)\bigr)+f\bigl(xf(x)-x\bigr) = 2x^3$ after the substitution: $(x,y) = (x,x)$
4. $f\bigl(-x-xf(x)\bigr) = -f(x)-x^3$ after the substitution: $(x,y) = (-x, x)$

Moreover I don't think this equation has any solutions...

Answer 1

You are correct. Use $f(-x)=-f(x)$ and substitute $y\to-y$ in the original equation to get:

$$f(x-yf(x))+f(y-xf(y))-f(x)-f(y) = 2xy^2$$

and here is the problem: LHS is symmetrical in $x$ and $y$, while RHS is not. Thus, substitute $x\to y$, $y\to x$ to get

$$f(x-yf(x))+f(y-xf(y))-f(x)-f(y) = 2x^2y.$$

We get that $2xy^2 = 2x^2y$, for all $x,y\in\mathbb R$, contradiction.