An explicit example of the blowup of an algebraic curve

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Consider the curve $x^3+y^4=0$ in $\mathbb{A}^2$, I am trying to retrieve the Dynkin diagram of $E_6$ by consecutive blowups at the origin, however I run into issues after performing just one of these morphisms.

The relations we get by blowing up at the origin are

$$x^3+y^4=0, \quad xv=uy$$ where $u,v$ are my homogeneous coordinates in $\mathbb{P}^1$. Let $\tilde{Y}$ denote the variety defined by these equations in $\mathbb{A}^2 \times \mathbb{P}^1$, I look for singularities in each chart.

When $u=1$ we have $xv=y$ and substituting into the first equation gives $0=x^3+x^4v^4=x^3(1+xv^4)$. The first factor tells us about our exceptional divisor while the second is nonsingular.

When $v=1$ we have $x=uy$ and substituting we have $0=u^3y^3+y^4=y^3(u^3+y)$, this second factor is also nonsingular, so I think I may have gone a bit wrong!

I was anticipating that I would have at least one singular thing in one of the charts so that I could keep blowing up and so on, could someone point out where my error is?