I think we dont need continuity. Let $g(x)=f(x)+2x$ $\implies$ $g(xy)=g(\frac{x^2+y^2}{2})$ $y =0$ gives $g(x)=c$ for all $x\ge 0$.$y=-x$ gives $g(x^2)=g(-x^2)$ so $g(x)=c$ for all $x\in R$ where $c$ is real constant. $f(x)=-2x+c$ which is a solution whatever $c$ is.

Let $P(x,y)$ be the assertion. $P(\sqrt{2x},0) \Rightarrow f(0)=f(x)+2x \Leftrightarrow f(x)=c-2x \quad \forall x\geqslant 0$ $P(x,-x) \Rightarrow f(-x^2)=f(x^2)+4x^2 \Rightarrow f(-x)=f(x)+4x=c-2x+4x=c-2(-x)$ Thus, $f(x)=c-2x \quad \forall x$. EDIT: sniped

Murad.Aghazade wrote: I think we dont need continuity. Yeah, I also wondered why continuity was given.

Let $P(x,y)$ be the assertion "$f(xy)=f\left(\frac{x^2+y^2}{2}\right)+(x-y)^2$". $P(x,0)$ ; $f(0)=f\left(\frac{x^2}{2}\right)+x^2$ $\rightarrow$ $f(x)=c-2x$ $\forall x\ge0$ where $c$ is a constant. $P(x,1)$ ; $f(x)=f\left(\frac{x^2+1}{2}\right)+(x-1)^2=c-x^2-1+x^2-2x+1=c-2x$ $\forall x\in\mathbb{R}$.

DerJan wrote: Murad.Aghazade wrote: I think we dont need continuity. Yeah, I also wondered why continuity was given. Everyone does...

continuity is not used

It is strange. Because it has been appeared here; https://artofproblemsolving.com/community/c6h1341597p7286065

$P(x,0)$ forces $f(0) - x^2 = f(\frac{x^2}{2})$. Parameterizing, for a nonnegative real $t$ we have $f(0) - 2t = f(t)$. Since $\frac{x^2+y^2}{2}$ is always positive, we can write $f(xy) = f(0) - 2xy$. Since $xy$ can vary over all reals, we have $f(x) = -2x + k$ for all $x$. This obviously works.