Consider this curve $f(x,y)=0$ given by


$$ f(x,y) := y^3 + y^2 x + x^4 =0.$$
Is it obvious that after a change of coordinates near the origin, this 
curve is equivalent to 


$$ \hat{y}^2 \hat{x} + \hat{x}^4 = 0 $$


I think, these are both $D\_5$ singularities. It seems like the change of 
coordinates that would achieve this is of the form


$$ x = \hat{x} - y + c\_2 y^2 + c\_3 y^3 + \ldots $$


where $x$ is an infinite power series. We can kill off the coefficients 
of $y^n$, for all $n$. This would give us a factor of $\hat{x}$, i.e we get something 
of the form
$$ f = \hat{x}\cdot g $$
And then we can make another change of coordinate, so that $g$ becomes
$$ g = \hat{y}^2 + \hat{x}^3.$$ Is there a simpler way to prove this? 
And aside from proving the power series converges, is there anything 
missing in the proof? Everything is over the complex numbers.



What is the simplest way to represent a $D_5$ singularity?
For classifying plane curves singularities, the "coordinate approach" is not always the better one\*. For your question, the general case is in table 1, page 3, C.T.C Wall article" "sextic curves and quartic surfaces with higher multiplicity". For general methods, I recommend, C.T.C Wall article: "Notes in the classification of singularities"


In general, the singularities $J\_{r,i}$ or $E\_{r,i}$ (the notation is not uniform) are given by $y^3+y^2x^r+x^{3r+i}$ with $r \geq 1$,and $i \geq 0$. In your case $r=i=1$ and the singularity is actually $D\_5$. You can recognized it because it has two branches: one smooth, and another one with an $A\_2$ singularity. Those branches separates after one blowing up. (see Table A, from the latest reference), and they are the factors that you see in your calculation. This "branch behavior" is the\* definition of the $D\_5$ singularity, and the normal form is deduced from it. A detailed discussion is in Barth's book in compact complex surfaces, page 79.


I hope it helps!


Psd: I don't see anything missing in your argument, but it is "simpler" by using $D\_5$'s resolution. 


\*to my knowledge/opinion


