Finding all $ f: \mathbb{R} \to \mathbb{R}$ such that $ f(x+f(y))=f(x)+2xy^2+y^2f(y) $

Question

$$ f: \mathbb{R} \to \mathbb{R}, f(x+f(y))=f(x) + 2xy^2 + y^2f(y) $$

How can we solve this problem? This is my try, but can't go more.

$$ P(x, y): f(x+f(y))=f(x) + 2xy^2 + y^2f(y) \\ P(0, 0): f(f(0))=f(0) \\ let \ f(0) = a \Rightarrow f(a)=a. \\ P(a, a): f(2a)=a+2a^3+a^3 \\ P(a, 0): f(2a)=a \\ \Rightarrow 3a^3=0 \Rightarrow a = 0, f(0) = 0. \\ \ \\ Assume) \ \exists \ t \ s.t. \ t \ \neq 0, \ f(t)=0. \\ P(t, t): 0 = 2t^3 \Rightarrow t = 0, \text{Contradiction.} \\ \therefore f(t)=0 \Leftrightarrow t = 0. \ \\ P(0, y): f(f(y))=y^2f(y). \\ P(f(x), x): f(2f(x))=f(f(x))+2f(x) \cdot x^2 + x^2f(x) = 4f(f(x))=4x^2\cdot f(x) \\ P(2f(x), x): f(3f(x))=f(2f(x))+4f(x)\cdot x^2 + x^2f(x) = 9f(f(x)) = 9x^2 \cdot f(x) \\ \cdot \\ \cdot \\ \cdot \\ f(nf(x))=n^2f(f(x))=n^2x^2f(x) \ for \ \forall n \in \mathbb{N}. \\ \ \\ P(-f(x), x): f(0)=f(-f(x))-2f(x) \cdot x^2 + x^2f(x) \Rightarrow f(-f(x)) = x^2f(x) = f(f(x)) \\ P(-2f(x), x): f(-f(x))=f(-2f(x))-4f(x)\cdot x^2 + x^2f(x) \Rightarrow f(-2f(x))=4x^2f(x)=4f(f(x)) \\ \cdot \\ \cdot \\ \cdot \\ f(mf(x))=m^2f(f(x))=m^2x^2f(x) \ for \ \forall m \in \mathbb{Z}. \\ $$

Answer 1

$$f(x+f(y))=f(x)+2xy^2+y^2f(y)$$

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• $(x,y)\equiv (0,0)$ $$f(f(0))=f(0)\implies f(c)=c\;,\;\;\; c=f(0).$$
• $(x,y)\equiv (0,c)$ $$f(f(c))=c+c^2f(c)\implies c^3=0\implies f(0)=c=0. $$

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• $(x,y)\equiv (0,x)$ $$f(f(x)=x^2f(x)\tag{1} $$
• $(x,y)\equiv (f(x),1)$ $$f(f(x)+f(1))=f(f(x))+2f(x)+f(1)\tag{2}$$
• $(x,y)\equiv (f(1),x)$ $$f(f(1)+f(x))=f(f(1))+2f(1)x^2+x^2f(x)\tag{3}$$
• $(x,y)\equiv (0,1)$ $$ f(f(1))=f(1)\tag{4}$$

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Adding $(1)$ and $(2)$, then subtracting $(3)$ and $(4)$ from the sum leads to, $$f(x)=f(1)x^2$$ Substituting this result in the original F.E, we have, $$f(1)^2=1\implies f(1)=\pm 1$$ $$\therefore\; \boxed{f(x)=\pm \;x^2} \;\;\forall x\in \mathbb{R} $$