Attempt so far:
The eigenvalue equation for $\frac{-d^2}{dx^2}$ is $y''(x) + \lambda y(x)=0$
If $\lambda =0$, $y(x) = Ax + B$ where A and B are constants.
The boundary conditions require that $B = -(A\pi +B) \implies $ and $y'(x) = A$ . Hence $A=B=0$, and $\lambda = 0$ is not an eigenvalue. (EDIT)
If $\lambda \neq 0$, then $-\lambda$ has two distinct square roots, ie $ u = \pm \sqrt{-\lambda}$.
Then the solution has the form $y \left( x \right) =A{{\rm e}^{ux}}+B{{\rm e}^{-ux}}$ $\implies y'(x) = u( A{{\rm e}^{ux}}-B{{\rm e}^{-ux})}$
$y(0) = -y(\pi) \implies A+B = -(A{{\rm e}^{u\pi}}+B{{\rm e}^{-u\pi}}) \implies A = -B e^{- \pi u} $ (1)
$y'(0) = -y'(\pi) \implies (A-B)u = -u(Ae^{u \pi} - B e^{-u \pi}) \implies A = B e^{- \pi u} $ (2)
(1) and (2) $\implies 2Be^{-\pi u}=0$ which means no solutions for $u$. Where have I gone wrong, and have I done previous parts correctly? Thanks in advance for any help.