Functional equation $f((xf(x))^2 + f(y))=-x^4 + y$

Question

Problem

Functional equation

Suppose $f\colon\mathbb{R}\to\mathbb{R}\quad$ $\forall x, y \in \mathbb{R}, f((xf(x))^2 + f(y))=-x^4 + y$

What I found :

Put $x=y=0,$ then $f(f(0))=0$

And put $x=f(0), y=0\quad$ I got $f(0)=0$ or $1$

Suppose $f(0)=1, $ then $f(1)=0 $ from $f(f(0))=0$

Put $x=1, y=0\quad$ (LHS) is $0$ but (RHS) is $-1$

Thus, $f(0)=0$

What should I do next?

Answer 1

1. Put $ x = f(0) $. Then you'll get $ f(f(y)) = y $
2. Put $f(x)$ instead of $x$. You'll get $f(f(x)^2f(f(x))^2 + f(y)) = f(f(x)^2x^2 + f(y)) = f((xf(x))^2 + f(y)) = -f(x)^4 + y$. But LHS is the same in the initial equation. Thus $ - x^4 + y = -f(x)^4 + y$ so $f(x)^4 = x^4$. Here you should suppose that there are two sets: $A = \{x | f(x) = x\}$ and $B = \{x | f(x) = -x\}$. You'll deal with the rest by yourself ;)