Behaviour of solution of autonomous ODE

I need to determine whether the solution to the ODE \begin{equation} y^\prime = -x\exp(\frac{y^{2\alpha}}{2}), \qquad \alpha \in (0,1) \end{equation} is such that $y = o(\frac{1}{\sqrt{x}})$ as $x \to 0^+$. I made the change of variable $z(x) = y(\sqrt{x})$. The ODE now reads \begin{equation} z^\prime = -\frac{1}{2}\exp(\frac{z^{2\alpha}}{2}) \end{equation} and I only have to check that $z = o(\frac{1}{x})$ as $x \to 0^+$. I was hoping to use some asymptotic expansion for the solution but I am having problems with the fact that $\alpha$ can be irrational.

Any help would be very appreciated, thanks in advance!

An autonomous equation is autonomous. Therefore, its solutions are related by time shifts: if $z=f(t)$ is a solution, then so is $z=f(t+C)$ for any constant $C$. In particular, there is nothing special about $x\to 0^+$. If one solution blows up going backwards in time, then they all do, and in exactly the same way, but at different times.
Let's see how they do it. Let $(x_0,z_0)$ be the initial condition, from which we move backwards in time. If the solution exists at time $x_1<x_0$, then $$\int_{z_0}^{z(x_1)} 2\exp(-z^{2\alpha}/2)\,dz = x_0-x_1 \tag1$$ The left side of (1) cannot exceed $\int_{z_0}^{\infty} 2\exp(-z^{2\alpha}/2)\,dz$. Therefore, a blow-up occurs at time $T<x_0$ such that $$x_0-T = \int_{z_0}^{\infty} 2\exp(-z^{2\alpha}/2)\,dz \tag2$$
So, depending on your initial condition, one of the following three things will happen as $x\to 0^+$:
• If $T<0$, then $z(x)$ is continuous at $x=0$. The blow-up is yet to occur.
• If $T>0$, then the solution blows up before reaching $x=0$. The question of its behavior as $x\to 0^+$ is moot.
• If $T=0$, we should consider the asymptotics of the integral in formula (2), plugging an arbitrary $x>0$ instead of $x_0$. This yields $$x = \int_{z(x)}^\infty 2\exp(-z^{2\alpha}/2)\,dz \tag3$$ For sufficiently large $z$ we have $2\exp(-z^{2\alpha}/2)\le z^{-10}$ (this is also true with other numbers in place of $10$). Therefore, for $x$ sufficiently close to $0$, $$x \le \int_{z(x)}^\infty z^{-10}\,dz = \frac{1}{9z(x)^9}\tag4$$ Rearrange (4) as $z(x)\le (9x)^{-1/9}$ and you are done.