North Macedonia

For $x=y=0$ we get $f(2a)=-a$ (where $a=f(0)$).

For $x=0$ and $y=2a$ we get $a-a=-a(2a{)}^k$, so $a=0$.

For $x=y=1$ we get $f(2b)=2b$ (where $b=f(1)$).

For $x=2b$ and $y=0$ we get $4b=(2b{)}^{k+1}$, so $b=0$ or $b=\frac{k\sqrt{2}}{2}$.

For $b=0$, $f(x)=0$ for every real number $x$, satisfies the equation.

For $k=2$ and $b=\frac{\sqrt{2}}{2}$, $f(x)=\frac{x^2}{\sqrt{2}}$, satisfies the equation.

For $k\ne 2$, for $x=1$ and $y=0$ we get $2f(b)=b$, i.e. $f(b)=\frac{b}{2}$.

For $x=y=b$, we get $f(b)=2b^k\frac{b}{2}$, i.e. $b^k=\frac{1}{2}$. But that is contradictory to $b^k=\frac{1}{2^{k-1}}$ (this is for $k\ne 2$).

Therefore, for $k=2$, $f(1)$ can be $0$ or $\frac{\sqrt{2}}{2}$, and for $k\ne 2$, $f(1)=0$.