
The first thing is to solve it numerically to vizualize how $x_\varepsilon$ varies in function of $\varepsilon$.
Above is the graph of :$\displaystyle\quad\frac{\ln(\varepsilon)}{\ln(x_\epsilon)}$ when $\varepsilon=10^{-a}\quad a\in[1,10]$
So we can consider that $\varepsilon\ll 1\implies \varepsilon=O(x^3)$ or similarly $x=O(\sqrt[3]\varepsilon)$
Now we can compare the development of $\displaystyle \sqrt{1+\frac{\varepsilon}{x^2}}$ and the one of $e^x$ since $\displaystyle \frac{\varepsilon}{x^2}=O(x)$.
$\displaystyle e^x=1+x+\frac 12 x^2+\frac 16 x^3+O(x^4)$
$\displaystyle 1+\sqrt{x^2+\varepsilon}=1+x\left(1+\frac 12 \frac{\varepsilon}{x^2} -\frac 18 \frac{\varepsilon^2}{x^4}+O(x^3)\right)=1+x+\underbrace{\frac 12 \frac{\varepsilon}{x}}_{O(x^2)} -\underbrace{\frac 18 \frac{\varepsilon^2}{x^3}}_{O(x^3)}+O(x^4)$
If we consider the development up to $O(x^2)$ we get :
$\displaystyle \frac 12 x^2\sim\frac 12\frac{\varepsilon}x\iff \bbox[5px,border:2px solid]{x\sim\sqrt[3]{\varepsilon}}\quad$ this is more precise than just $x=O(\sqrt[3]{\varepsilon})$.
If we consider now the development up to $O(x^3) $ we get :
$\displaystyle \frac 12 x^2+\frac 16x^3=\frac 12\frac{\varepsilon}x-\frac 18\frac{\varepsilon^2}{x^3}+o(\varepsilon)$
Let's have $x^3=\varepsilon+u\quad$ with $u=o(\varepsilon)\quad$
(we could have instead $x=\sqrt[3]{\varepsilon}+u$, but it complicates the calculation)
$\begin{array}{l}\require{cancel}
12x^5+4x^6=12x^2\varepsilon-3\varepsilon^2+o(\varepsilon^2)\\
12x^2(x^3-\varepsilon)+4(u+\varepsilon)^2+3\varepsilon^2=o(\varepsilon^2)\\
12x^2u+\cancel{4u^2}+\cancel{8u\varepsilon}+7\varepsilon^2=o(\varepsilon^2) & u^2,u\varepsilon=o(\varepsilon^2)\text{ so we cancel them }
\end{array}$
In the same way $12x^2u\sim 12\varepsilon^\frac 23u$ other terms are negligible.
So we get $\quad\displaystyle 12\varepsilon^\frac 23u\sim-7\varepsilon^2\iff u\sim -\frac 7{12}\varepsilon^\frac 43$
Now reporting in $x^3=\varepsilon+u\quad$ we get $\quad \bbox[5px,border:2px solid]{x=\varepsilon^\frac 13-\frac 7{36}\varepsilon^\frac 23+o(\varepsilon^\frac 23)}$
We could continue the development, but it becomes delicate to know which terms we can keep and which ones we can dismiss.