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 $k \neq 2$, 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 $k \neq 2$).

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