(Possible) Misunderstanding of perturbation method for finding solution of polynomial equation? This is the strange moment that I get when I solve this equation:
$$ \frac{w^4}{4} - \frac{w^3}{3} = \varepsilon, $$
where $\varepsilon$ is a small parameter. If I plot the graph $ w \mapsto \frac{w^4}{4} - \frac{w^3}{3}$, I can see that there are two real solutions to this equation. I am interested in the solution that is close to $0$ and I want to find its asymptotics in terms of $\varepsilon$. 
(I have to warn that I have never used perturbation techniques before, so forgive me if I am using wrong terminology or if I do stupid mistakes)
It seems that finding solution in form $\hat{w}(\varepsilon) = a_0 + a_1 \varepsilon + \dots$ doesn't work here. I am using this text as a reference and this is an expected result according to it . My equation looks similar to Example $5.3$ on page 8, so I've decided to check substitution $w = \varepsilon^{p} \cdot y$ using considerations from this example.
This lead me to the substitution $\hat{w} = \varepsilon^{\frac{1}{3}} \cdot y(\varepsilon), \; y(0) \neq 0$, and I got an equation:
$$ \frac{\varepsilon^{\frac{4}{3}}\cdot y^4}{4} - \frac{\varepsilon \cdot y^3}{3} = \varepsilon .$$
It seems to be the right thing to divide both sides by $\varepsilon$, but then I get
$$ \frac{\varepsilon^{\frac{1}{3}}\cdot y^4}{4} - \frac{y^3}{3} = 1 .$$
This looks like that now solution needs singular perturbation methods. But the original equation didn't seem to need them! This is the moment where I am kind of lost, because I expected to get an equation similar to the equation from Example $5.3$.
So I have two questions:
1. Did I use the wrong method?
2. If I used the right method, why did it lead from regular perturbation problem to singular perturbation problem? 
Thanks in advance for your help!
 A: From your original equation, you can do a regular perturbation expansion $w=w_0+\epsilon w_1+\epsilon^2w_2+\ldots$ to get, at $O(1)$,
$$w_0^4-\frac{4}{3}w_0^3=0,$$
which has three solutions $w_0=0$ and one solution $w_0=4/3$. At $O(\epsilon)$, you get $w_0^3w_1-w_0^2w_1=1$, which is not consistent for the roots $w_0=0$, so consider $w_0=4/3$ to give,
$$ \left(\frac{4}{3}\right)^3w_1-\left(\frac{4}{3}\right)^3w_1=1$$
so $w_1=27/16$. Then, the $O(\epsilon^2)$ equation gives $w_2=1215/2048$. So the root near $4/3$ is
$$w=\frac{4}{3}+\epsilon\frac{27}{16}+\epsilon^2\frac{1215}{2048}+O\left(\epsilon^3\right).$$
For the roots near zero, our naive expansion $w=w_0+\epsilon w_1+\epsilon^2w_2+\ldots$ failed, so, as you did, set $w=\epsilon^py$, from which you determine $p=1/3$, and so
$$ \frac{\epsilon^{1/3}y^4}{4}-\frac{y^3}{3}=1.$$
Now, write $y$ as a series in one-third powers in $\epsilon$ (so that you can deal with the $\epsilon^{1/3}$ in the equation above), $ y=y_0+\epsilon^{1/3}y_1+\epsilon^{2/3}y_2$. The $O(1)$ equation is
$$-\frac{y_0^3}{3}=1,$$
so $y_0=-3^{1/3}$ is the only real solution. The next equation is
$$\frac{y_0^4}{4}-y_0^2y_1=0,$$
so $y_1=9^{1/3}/4=(3/8)^{2/3}$. The $O(\epsilon^{2/3})$ equation is 
$$y_0^3y_1-\frac{4y_0^2y_1}{3}+2y_0^2y_1^2=0,$$
with solution $$y_2=\frac{3^2\left(3^{1/3}-2\right)}{2^5}.$$
So, the root near zero is
$$w=-\left(3\epsilon\right)^{1/3}+\left(\frac{3\epsilon}{8}\right)^{2/3}+\frac{3^2\left(3^{1/3}-2\right)}{2^5}\epsilon+O\left(\epsilon^{4/3}\right).$$
