Let $\ Y_i \sim f_{\theta}(y) = \frac{\theta}{2\sqrt{y}}\cdot e^{-\theta \sqrt{y}},\ \ \ i = 1, \dots,n, \ \ y \ge 0,\ \ \theta \ge 0 $
I'm trying to find out if a gamma prior $\ (\alpha,\beta) $ will be a conjugate to the data
$$\ P(\theta |Y) = \frac{P(Y|\theta )\cdot P(\theta) }{P(Y)} \propto P(Y|\theta)P(\theta) = \prod_{i=1}^n \frac{1}{2\sqrt{y_i}} \cdot e^{-\theta \sqrt{y_i}} \cdot \frac{\theta^{\alpha-1} e^{-\beta\theta} \beta^{\alpha}\\}{\Gamma(\alpha)} \\ \propto \frac{e^{-\theta\sum\sqrt{y_i}}}{2^n} \cdot \theta^{\alpha-1} e^{-\beta\theta} \beta^{\alpha} \prod \sqrt{y_i}^{-1} $$
I can't really see how to proceed from here to something similar to the gamma density form (if possible) ?
Your likelihood is
$$p(\mathbf{y}|\theta)\propto \theta^n e^{-\theta\sum\sqrt{ y}}$$
for the sake of simplicity I wasted any quantity not depending on $\theta$
The prior is
$$\pi(\theta)\propto \theta^{a-1}e^{-b\theta}$$
thus to derive the posterior, simply multiply $\text{likelihood}\times\text{prior}$ obtaining
$$\pi(\theta|\mathbf{y})\propto \theta^{a+n-1}\cdot e^{-\theta(b+\sum\sqrt{ y})}$$
and we immediately recognize the kernel of a Gamma distribution
$$\pi(\theta|\mathbf{y})\sim\Gamma\left(a+n;b+\sum\sqrt{ y}\right)$$
If you want to get the exact posterior density simply normalize obtaining
$$\pi(\theta|\mathbf{y})=\frac{(b+\sum\sqrt{ y})^{a+n}}{\Gamma(a+n)}\theta^{a+n-1}\cdot e^{-\theta(b+\sum\sqrt{ y})}$$