We have x^2+xy+ y^2= 0\cdots(1)
if x is zero the y is zero then x+y = 0 which is not possible as x+y is in denominator of the resultnat expression
Let y=x\omega
putting in (1) we get
\omega ^2 + \omega + 1=0\cdots(2)
Hence \omega^3=1\cdots(2)
Now x+y = x + x\omega = x(1+\omega) = x(-\omega^2)= - x\omega^2 using(1)
hence \frac{x}{x+y} = \frac{x}{- x\omega^2} = - \omega\cdots(3)
also \frac{y}{x+y} = \frac{x\omega}{- x\omega^2} = - \frac{1}{\omega} = \omega^2\cdots(4)
hence (\frac{x}{x+y})^{2023} + (\frac{y}{x+y})^{2023}) = \omega^{2023} + (\omega^2)^{2023}
= - \omega^{2023} - \omega^{4036}
= - \omega^{3 * 677 + 1} - \omega^{3 * 1354 + 2 }
= - \omega - \omega^2 using (3)
= 1 using (2)