Method 1: Residues
Consider the contour integral
$$\oint_C dz \frac{e^{i k z}}{(1+z^2)^2}$$
where $C$ is a semicircle in the upper half plane; here, $k>0$. Then by the residue theorem and Jordan's lemma:
$$\begin{align}\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{(1+x^2)^2} &= i 2 \pi \operatorname*{Res}_{z=i} \frac{e^{i k z}}{(1+z^2)^2}\\ &= i 2 \pi \left [\frac{d}{dz} \frac{e^{i k z}}{(z+i)^2} \right ]_{z=i}\\ &= i 2 \pi \left [\frac{i k\, e^{i k z}}{(z+i)^2} - \frac{2 e^{i k z}}{(z+i)^3} \right ]_{z=i}\\ &= \frac{\pi}{2} (k+1) e^{-k}\end{align}$$
For $k \lt 0$, $C$ is a semicircle in the lower half plane; for the same reasons as above, we have:
$$\begin{align}\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{(1+x^2)^2} &= -i 2 \pi \operatorname*{Res}_{z=-i} \frac{e^{i k z}}{(1+z^2)^2}\\ &= -i 2 \pi \left [\frac{d}{dz} \frac{e^{i k z}}{(z-i)^2} \right ]_{z=-i}\\ &= -i 2 \pi \left [\frac{i k\, e^{i k z}}{(z-i)^2} - \frac{2 e^{i k z}}{(z-i)^3} \right ]_{z=-i}\\ &= \frac{\pi}{2} (-k+1) e^{k}\end{align}$$
Therefore,
$$\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{(1+x^2)^2} = \frac{\pi}{2} (1+|k|) e^{-|k|}$$
Method 2: Convolution
Knowing that the FT of $1/(1+x^2)$ is $\pi \, e^{-|k|}$, we may use the convolution theorem to deduce that
$$\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{(1+x^2)^2} = \frac{\pi^2}{2 \pi} \int_{-\infty}^{\infty} dk' e^{-|k'|} \, e^{-|k-k'|}$$
Again, the way we evaluate the integral on the RHS depends on the sign of $k$. For $k \gt 0$, this integral is
$$\frac{\pi}{2} \int_{-\infty}^0 dk' \, e^{k'} \, e^{-(k-k')} + \frac{\pi}{2} \int_{0}^k dk' \, e^{-k'} \, e^{-(k-k')}+ \frac{\pi}{2} \int_{k}^{\infty} dk' \, e^{-k'} \, e^{k-k'}$$
For $k \lt 0$, on the other hand, the integral is
$$\frac{\pi}{2} \int_{-\infty}^k dk' \, e^{k'} \, e^{-(k-k')} + \frac{\pi}{2} \int_{k}^0 dk' \, e^{k'} \, e^{k-k'}+ \frac{\pi}{2} \int_{0}^{\infty} dk' \, e^{-k'} \, e^{k-k'}$$
Evaluation of the above integrals reproduces the result derived above.
As already stated it depends upon how you really define the Fourier transform.
If you define it as an improper integral the question becomes non-trivial (otherwise it is just Plancherel). For $n\geq 1$ let
$$ f_n(x) = f(x)\; {\rm \bf 1}_{[-n,n]}(x) = \frac{1}{\sqrt{1+x^2}} {\rm \bf 1}_{[-n,n]}(x) $$
where ${\rm \bf 1}_{[-n,n]}$ is the indicator function for the interval $[-n,n]$. Then $f_n$ is in both $L^1\cap L^2$ and the Fourier transform
$$ \hat{f}_n(t) = \int_{-n}^n \frac{e^{itx}}{\sqrt{1+x^2}} dx $$
converges for every $t\neq 0$. This follows from partial integration:
$$ \hat{f}_n(t) = \left[ \frac{e^{itx}}{it\sqrt{1+x^2}} \right]_{-n}^n + \int_{-n}^n \frac{\sin(tx)}{t} \frac{x}{\sqrt{1+x^2}^3} dx . $$
As $n\rightarrow \infty$, the first term goes to zero and the latter converges absolutely.
Thus, you may define $\hat{f}(t)=\lim_n \hat{f}_n(t)$ for every $t\neq 0$ and the question is then if $\hat{f}\in L^2$?
The answer is yes: First note, that $f_n \rightarrow f$ in $L^2$ so $(f_n)_{n\geq 1}$ is Cauchy in $L^2$.
By Plancherel $\|\hat{f}_n-\hat{f}_m\|_2 = \|f_n-f_m\|_2$, so $(\hat{f}_n)_{n\geq 1}$ is also Cauchy in $L^2$, whence converges in $L^2$ to some function $g$.
Any convergent sequence in $L^2$ has, however, a subsequence that converges a.e. and we already know that for $t\neq 0$ (i.e. a.e.) the limit is $\hat{f}(t)$, so $\hat{f}=g$ (mod 0) is in $L^2$
Best Answer
We begin by folding the integral and enforcing the substitution $x\mapsto x^2$. This reveals that
$$\begin{align} \int_{-\infty}^\infty \frac{e^{-|x|}}{\sqrt{|x|}}e^{-ixy}\,dx&=2\int_0^\infty \frac{e^{-x}}{\sqrt x}\,\cos(xy)\,dx\\\\ &=2\text{Re}\left(\int_0^\infty \frac{e^{-(1-iy)x}}{\sqrt x}\,dx\right)\\\\ &=4\text{Re}\left(\int_0^\infty e^{-\left(\sqrt{1-iy}\,x\right)^2}\,dx\right)\\\\ \end{align}$$
Next, we enforce the substitution $\sqrt{1-iy}\,x\mapsto x$ to obtain
$$\int_{-\infty}^\infty \frac{e^{-|x|}}{\sqrt{|x|}}e^{-ixy}\,dx=4\text{Re}\left(\frac1{\sqrt{1-iy}}\int_0^{\infty/\sqrt{1-iy}} e^{-x^2}\,dx\right)$$
Invoking Cauchy's Integral Theorem, we can deform the contour back onto the real line to find
$$\begin{align} \int_{-\infty}^\infty \frac{e^{-|x|}}{\sqrt{|x|}}e^{-ixy}\,dx&=4\text{Re}\left(\frac1{\sqrt{1-iy}}\int_0^{\infty} e^{-x^2}\,dx\right)\\\\ &=2\sqrt{\pi}\text{Re}\left(\frac1{\sqrt{1-iy}}\right)\\\\ &=\frac{2\sqrt{\pi}}{\sqrt{1+y^2}}\text{Re}\left(\sqrt{1+iy}\right)\\\\ &=\frac{2\sqrt{\pi}}{\sqrt{1+y^2}}\text{Re}\left(\sqrt{\sqrt{1+y^2}e^{i\arctan(y)}}\right)\\\\ &=\frac{2\sqrt{\pi}}{\sqrt{1+y^2}} \sqrt{\sqrt{1+y^2}}\cos(\arctan(y)/2)\\\\ &=\frac{2\sqrt{\pi}}{\sqrt{1+y^2}} \sqrt{\sqrt{1+y^2}}\sqrt{\frac{1+\cos(\arctan(y))}{2}}\\\\ &=\sqrt{2\pi}\frac{\sqrt{\sqrt{1+y^2}\left(1+\frac1{\sqrt{1+y^2}}\right)}}{\sqrt{1+y^2}}\\\\ &=\sqrt{2\pi}\sqrt{\frac{\sqrt{1+y^2}+1}{1+y^2}} \end{align}$$