$ y=x $ $ f(f(x)^2+f(x^2))=0 $ $ y=-x $ $ f(f(x)^2+f(x^2)=2x(f(x-f(-x))=0 $ $ f(x-f(-x))=0 $ $ f(x)=0 $ and $ f(x)=-x $

Feridimo wrote: $ y=x $ $ f(f(x)^2+f(x^2))=0 $ $ y=-x $ $ f(f(x)^2+f(x^2)=2x(f(x-f(-x))=0 $ $ f(x-f(-x))=0 $ $ f(x)=0 $ and $ f(x)=-x $ You can't conclude that; first you don't have that $f (0)=0$ and $f (f(x)^2+f (x^2)) $ don't gives all the values to conclude $f (x)=0$

mathisreal wrote: Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $f(f(x)^2+f(y^2))=(x-y)f(x-f(y))$ I suppose that missing domain of functional equation is "$\forall x,y\in\mathbb R$". If so : Let $P(x,y)$ be the assertion $f(f(x)^2+f(y^2))=(x-y)f(x-f(y))$ Subtracting $P(x,y)$ from $P(x,-y)$, we get New assertion $Q(x,y)$ : $(x-y)f(x-f(y))=(x+y)f(x-f(-y))$ 1) If $\exists u\ne 0$ such that $f(u)=0$, then $f\equiv 0$ Let $u\ne 0$ such that $f(u)=0$ $Q(-u,-u)$ $\implies$ $f(-u)=0$ $Q(x,u)$ $\implies$ $\boxed{\text{S1 : }f(x)=0\quad\forall x}$ which indeed is a solution. 2) If $f(x)=0$ implies $x=0$ then $f(x)=-x$ $\forall x$ $Q(-x,-x)$ $\implies$ $f(-x-f(x))=0$ $\forall x\ne 0$ And so $f(x)=-x$ $\quad\forall x\ne 0$ Then $P(1,1)$ $\implies$ $f(0)=0$ And so $\boxed{\text{S2 : }f(x)=-x\quad\forall x}$ which indeed is a solution.

pco wrote: mathisreal wrote: Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $f(f(x)^2+f(y^2))=(x-y)f(x-f(y))$ I suppose that missing domain of functional equation is "$\forall x,y\in\mathbb R$". If so : Let $P(x,y)$ be the assertion $f(f(x)^2+f(y^2))=(x-y)f(x-f(y))$ Subtracting $P(x,y)$ from $P(x,-y)$, we get New assertion $Q(x,y)$ : $(x-y)f(x-f(y))=(x+y)f(x-f(-y))$ 1) If $\exists u\ne 0$ such that $f(u)=0$, then $f\equiv 0$ Let $u\ne 0$ such that $f(u)=0$ $Q(-u,-u)$ $\implies$ $f(-u)=0$ $Q(x,u)$ $\implies$ $\boxed{\text{S1 : }f(x)=0\quad\forall x}$ which indeed is a solution. 2) If $f(x)=0$ implies $x=0$ then $f(x)=-x$ $\forall x$ $Q(-x,-x)$ $\implies$ $f(-x-f(x))=0$ $\forall x\ne 0$ And so $f(x)=-x$ $\quad\forall x\ne 0$ Then $P(1,1)$ $\implies$ $f(0)=0$ And so $\boxed{\text{S2 : }f(x)=-x\quad\forall x}$ which indeed is a solution. Thanks https://artofproblemsolving.com/community/c6h1966517_functional_equations

mathisreal wrote: Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $f(f(x)^2+f(y^2))=(x-y)f(x-f(y))$ $\color{blue}\boxed{\textbf{Answer:}f(x)\equiv 0, -x}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $$f(f(x)^2+f(y^2))=(x-y)f(x-f(y))...(\alpha)$$In $(\alpha) y=x:$ $$\Rightarrow f(f(x)^2+f(x^2))=0$$$$\Rightarrow \exists a / f(a)=0$$In $(\alpha) y=-y:$ $$\Rightarrow f(f(x)^2+f(y^2))=(x+y)f(x-f(-y))$$By $(\alpha):$ $$\Rightarrow (x-y)f(x-f(y))=(x+y)f(x-f(-y))...(I)$$In $(I) y=a:$ $$\Rightarrow (x-a)f(x)=(x+a)f(x)$$$$\Rightarrow f(x)\equiv 0 \text{ or }x-a=x+a$$$$\Rightarrow f(x)\equiv 0 \text{ is a solution}...(1)$$If $x-a=x+a:$ $$\Rightarrow a=0$$$$\Rightarrow f(0)=0, \text{ and }f\text{ is inyective at 0}...(II)$$In $(I) x=-x, y=-x, x\neq 0:$ $$\Rightarrow 0=-2xf(-x-f(x))$$$$\Rightarrow f(-x-f(x))=0$$By $(II):$ $$\Rightarrow f(x)\equiv -x...(2)$$By $(I)$ and $(II):$ $$\Rightarrow \boxed{f(x)\equiv 0,-x \textbf{ are the only solutions}_\blacksquare}$$$\color{blue}\rule{24cm}{0.3pt}$

Denote $P(x,y)$ as the assertion of the given F.E. $P(x,x)$ gives that $0$ is in the domain of $f$, so let $f(c)=0$ for some $c \in \mathbb R$. Now from $P(x,-x)$ we get $0=f(f(x)^2+f(x^2))=2x f(x-f(-x))$ so if $x \ne 0$ then $f(x-f(-x))=0$, now suppose FTSOC that $c \ne 0$, then from here we get $f(-c)=0$, therefore $P(x,c)-P(x,-c)$ give that $(x-c)f(x)=(x+c)f(x)$ which means that $f(x)=0$ for all reals $x$ which works. Now if $f(c)=0$ forced $c=0$ then we have $f(0)=0$ and also $x-f(-x)=0$ which means that $f(x)=-x$ for all reals $x$ which indeed works as well, thus we are done .