Eigenfunction of the Fourier transform

Hint: let $z\in \mathbb C$. The zeroes of $f(z)=\exp(z)+\exp(-z)$ are all complex numbers $z_k=a+ib_k$ s.t.

$$a=0$$ $$b_k=\frac{\pi}{2}+k\pi,~~k\in\mathbb Z.$$

The integral $\int_{-\infty}^{\infty}\frac{\exp(-i\tilde{\omega}y)}{f(y)}dy,$ with $\tilde{\omega}:=\omega\sqrt{2}{\pi}$ can be computed using the residue theorem and an appropriate limiting procedure ($R\rightarrow \infty$, $K\rightarrow \infty$) on

$$\int_{\gamma}\frac{\exp(-i\tilde{\omega}z)}{f(z)}=Res(f(z),z_0)=\frac{\exp((-i\tilde{\omega}i\frac{\pi}{2})}{2exp(i\frac{\pi}{2})}=\frac{1}{2}\exp(\tilde{\omega}\frac{\pi}{2}),$$

with $\gamma=\gamma_1+\gamma_2$, and $\gamma_1(t)=-R(1-t)+tR$, $\gamma_2(t)=K\exp(i\pi t)$, $z_0=i\frac{\pi}{2}$. Here $K>1$ and $t\in[0,1]$.

You are just about there. One trick that I find helps over this hump is to recognize that your sum may be rewritten as

$$ \frac{2}{\pi} \sum_{k=-\infty}^{\infty} (-1)^k \frac{ 2k+1 }{ (2k+1)^2 + (w \sqrt{\frac{2}{\pi}} )^2}$$

Now you can apply the residue theorem to this infinite sum. I will state the following result without (much) proof: the following convergent sum satisfies

$$\sum_{k=-\infty}^{\infty} (-1)^k \, f(k) = -\sum_m \text{Res}_{z=z_m} \pi \csc{(\pi z)} f(z)$$

where $z_m$ are the poles of $f$ that are not real integers. You may prove this by integrating the function $\pi \csc{(\pi z)} f(z)$ along a rectangular contour about the real line interval $x \in [-N,N]$, and taking the limits as $N \to \infty$. Now in this case,

$$f(z) = \frac{2}{\pi}\frac{ 2k+1 }{ (2k+1)^2 + (w \sqrt{\frac{2}{\pi}} )^2}$$

Your poles are at $z_{\pm} = -(1/2) \pm i w \sqrt{\frac{1}{2 \pi}} $. Now simply take these poles and plug into the above residue formula:

$$\frac{2}{\pi}\sum_{k=-\infty}^{\infty} (-1)^k \frac{ 2k+1 }{ (2k+1)^2 + (w \sqrt{\frac{2}{\pi}} )^2} = - \left [\csc{\left(-\frac{\pi}{2}+i w \sqrt{\frac{\pi}{2}}\right)} + \csc{\left(-\frac{\pi}{2}-i w \sqrt{\frac{\pi}{2}}\right)} \right ]$$

After simplification, I get as your FT:

$$\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} dx \, \frac{e^{-i w x}}{\cosh{[x \sqrt{(\pi/2)}]}} = \frac{2}{\cosh{[w \sqrt{(\pi/2)}]}}$$

which would certainly serve as an eigenfunction of the FT.