Here is a related problem. First solve the differential equation
$ y(x)=c_1\,\sin \left( \sqrt {\lambda}x \right) + c_2\,\cos\left( \sqrt {\lambda}x \right) .$ Applying the boundary conditions to the solution results in the two equations
$ y(\pi)=y(-\pi) \implies 2{ c_1}\,\sin \left( \sqrt {\lambda}\pi \right)= 0 \rightarrow (1) $
$y'(\pi)=y'(-\pi)\implies 2{c_2}\sqrt{\lambda}\sin \left( \sqrt {\lambda}\pi \right) = 0 \rightarrow (2), $
where $c_1$ and $c_2$ are arbitrary constants. From $(1)$, since $c_1$ is an arbitrary constant, then we have
$ \sin\left( \sqrt {\lambda}\pi \right) = 0 \implies \sqrt {\lambda}\pi = n\pi \implies \lambda = n^2. $
Eq. $(2)$ gives the same eigenvalues with the other eigenvalue $\lambda=0$, since $c_2$ is an arbitrary constant. So, can you find the eigenvectors now and answer the question?
Note: This problem is known as "The Sturm-Liouville Eigenvalue Problems". One of the properties of this kind of problems is that "for each eigenvalue $\lambda_n$ there exists an eigenfunction $\phi_n$.