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My work : first take $x=y=0$ we get $f(0)=f(0)=c$ , then take $x=x$ and $y=-x$ implies $f(x^{2}-xf(x))=f(0)$ now as there is another function inside the parenthesis I'm asuming there are some entries for which $x^{2}-xf(x)=0$ then we get $f(x)=x$. but I'm little confused about that assumption. If anyone can just fix my mistakes and solve that problem it'll be a great help for me.

Aryaman Maithani
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Toneri Otsutsuki
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    f(x) = x isn't a solution as then we get x^2 + yx = x+y for all x in R; also f(x) is not necessarily injective so it may not have an inverse so you can't say x^2 - xf(x) = 0 from f(x^2 - xf(x)) = f(0) – Ben Feb 26 '21 at 12:50
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    Putting $y=0$ gives you $f(x^2)=f(x)$ for all $x$. And putting $x=0$ gives $f(yf(0))=f(y)$ for all $y$. – TonyK Feb 26 '21 at 12:58
  • I don't have time for an answer now but here are some of my thoughts. $f(x) = c$ is obviously a solution. I think these may be the only solutions.

    If there exists $a$ such that $f(a^2) = 0$ then $f(x) = 0$ for all x as $f(a^2) = f(a + y)$. Letting y vary shows that $f(x) = f(a^2)$ for all x so all non-constant solutions must have $f(x) \lt 0$ for all x or $f(x) \gt 0$ for all x

    f(x) can't be injective because, as you said, then $f(x) = x$ but that isn't a solution

     – Ben Feb 26 '21 at 13:24
  • Are you sure your functional equation is $f(x^{2}+yf(x))=f(x+y)$. If your functional equation is $f(x^{2}+yf(x))=xf(x+y)$ which is an problem from INMO 2015 then its solution can be found on AoPS.http://www.artofproblemsolving.com/community/c6h623452p3730730 – user6262 Feb 26 '21 at 13:50
  • yeah, I've taken it from INMO for a contest but forgot to put the x on RHS... because of this it had turned into another problem...that's why I'm looking for is there any solution of this FE rn... – Toneri Otsutsuki Feb 26 '21 at 14:02
  • You might have told us, Toneri! – TonyK Feb 26 '21 at 14:32
  • I did't cause it's still a valid ques @TonyK – Toneri Otsutsuki Feb 26 '21 at 15:26
  • Well that's a downvote from me then. Everybody thought that it was a valid Olympiad question or something, with an interesting answer. – TonyK Feb 26 '21 at 15:28
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    @TonyK It has got an interesting answer. $f(x)$ is an constant function. – user6262 Feb 26 '21 at 16:36
  • @user1055: srsly? – TonyK Feb 26 '21 at 22:15
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    @TonyK Do you think it is a joke – user6262 Feb 27 '21 at 05:21

2 Answers2

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As observed in the comments, $f(x^2)=f(x)$ for all $x$ so that $f(x)=f(-x)$ for all $x$. Thus, $f(x^2+yf(x))=f(x^2+yf(-x))$ for all $x$, and $f(x+y)=f(-x+y)$, for all $x,y$. This implies that $f$ is constant.

Aravind
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Let $x=x$ and $y=0$ $ \implies f(x^2)=f(x)$

Let $x=-x$ $\implies f(-x)=f(x)$ i.e $f(x) $ is an even function

$f(x^2+yf(x))=f(x^2+yf(-x)$ since $f(x)=f(-x)$

$\implies f(x+y)=f(-x+y)$

If we let $y=-x$ then $f(0)=f(-2x)$ which implies that $f(x)=f(0)$. We are not given value of $f(0)$ nor we are able to find it. Since $f(0)$ is constant which means that $f(x)$ is an constant function.

user6262
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