smc

We can write $$\begin{array}{rl}& x=\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)=e^{\frac{2\pi }{3}i},\text{ and}\\ & y=\cos \left(-\frac{2\pi }{3}\right)+i\sin \left(-\frac{2\pi }{3}\right)=e^{-\frac{2\pi }{3}i},\end{array}$$ which gives $$\begin{array}{rl}& x^k=e^{\frac{2\pi k}{3}i}=\cos \left(\frac{2\pi k}{3}\right)+i\sin \left(\frac{2\pi k}{3}\right),\text{ and}\\ & y^k=e^{-\frac{2\pi k}{3}i}=\cos \left(-\frac{2\pi k}{3}\right)+i\sin \left(-\frac{2\pi k}{3}\right)=\cos \left(\frac{2\pi k}{3}\right)-i\sin \left(\frac{2\pi k}{3}\right),\end{array}$$ using the fact that $\cos$ is an even function and $\sin$ is an odd function. Accordingly, $$x^k+y^k=2\cos \left(\frac{2\pi k}{3}\right),$$ and upon substituting the values $k=5,7,9,11,13$ from the answer choices, we find that $x^k+y^k=-1$ for all such values except $k=9$, where $x^9+y^9=2\cos (6\pi )=2\ne -1$. Thus the answer is $\boxed{x^9+y^9=-1}$.