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Find all function $f:\mathbb{R}\rightarrow \mathbb{R}$ so that $f(xf(y))+f(yf(x)=2xy$.
By putting $x=y=0$ we get $f(0)=0$ and by putting $x=y=1$ we get $f(f(1))=1$. Let $y=f(1)\Rightarrow f(x)+f(f(x)f(1))=2x$, which tells us that $f$ is an injective function. The only solutions I came up with so far are $f(x)=x$ and $f(x)=-x$.
Putting $x=1$ in $f(x)+f(f(x)f(1))=2xf(1)$ we get that $f(1)=\pm 1$
Let $f(1)=1$.
$x\rightarrow xf(x), y\rightarrow \frac{1}{x} \Rightarrow f(xf(x)f(\frac{1}{x}))+f(x)=2f(x) \rightarrow f(x)f(\frac{1}{x})=1$.
Putting now $x\rightarrow \frac{1}{x}, y \rightarrow 1$ we get $\frac{1}{f(x)}+ \frac{1}{f(f(x))}=\frac{2}{x}$. Using $f(f(x))=2x-f(x)$ we get $(f(x)-x)^2=0$, and there we get one of our two solutions, $f(x)=x$. We get $f(x)=-x$ as the other solution by putting $f(1)=-1$.
This is all I have found so far:
Just keep feeding the snake with its own tail. As you noted $f(f(1))=1$. So with $x=y=f(1)$ we then see that $1=f(f(1))=(f(1))^2$ so that $f(1)=\pm 1$. As you almost correctly noted, we have for $y=f(1)$ that $$ f(x)+f(f(x)f(1))=2xf(1) $$ which indeed shows that $f$ is injective. With that observation, the comment by Pp.. works to conclude that $f$ is odd since $f(xf(x))=x^2=f(-xf(-x))$ then implies $f(x)=-f(-x)$.
Now, if we knew that $f$ was a polynomial it would be easy to conlude that $f(xf(x))=x^2$ implies $\deg f(\deg f+1)=2$ so $f$ has to have degree 1 in that case. Since $f(0)=0$ and $f(1)=\pm 1$ that would then imply $f(x)=\pm x$.
But if $f$ is not a polynomial, all I can tell this far is that $f$ is an injective, odd function with $f(0)=0$ and $f(1)=\pm 1$ and a bucnh of functional equations attached to it. Maybe someone else can elaborate further ...
