Given the SL operator $L=-\frac{\mathrm{d}^2}{\mathrm{d}x^2}$ and the boundary conditions
$f'(0)=0,\ f'(1)+f(1)=0$
I want to compute eigenvalues and eigenvectors.
Since we have $f''+zf=0$ the solution is given by
$f=c_1\sin(\sqrt{z}x)+c_2\cos(\sqrt{z}x)$
The first boundary condition $f'(0)=0$ gives $c_1=0$ and the second one
$c_2\left ( \cos(\sqrt{z})-\sin(\sqrt{z})\sqrt{z}\right )=0$ so we need
$z=\mathrm{cotan}^2(\sqrt{z})$
I am not sure how to go from here.
That is all that there is to this problem. Perhaps you should add some remark why you exclude negative $z$ resp. how they are still included in the formula.
Then you can try to find quantitative properties of the eigenvalues, for large $z$ these will be close to the poles $\sqrt{z}\approx k\pi$ of the co-tangent. Then with $\sqrt{z}=k\pi+u_1$ $$ (k\pi+u_1)=\cot(k\pi+u_1)=\cot(u_1) $$ has the approximate solution $$ u_1\approx\tan(u_1)=\frac1{k\pi+u_1}\approx\frac1{k\pi}-\frac{u_1}{k^2\pi^2} $$ so that $$ u_1\approx\frac{k\pi}{1+k^2\pi^2}. $$
Numerical computation confirms that formula