Use the alternative affine coordinates
$$
t=-\frac{x}{y},\qquad w=-\frac1y,
$$
so that $x=t/w$ and $y=-1/w$. The equation becomes
$$
w=t^3+atw^2+bw^3.
$$
Recursive coefficient comparison gives a unique series $w(t)\in t^3k[[t]]$. Since $v(w(t))=3v(t)$ and hence $v(x)=-2v(t)$, $v(y)=-3v(t)$, this gives the <formal coordinates on a short Weierstrass curve> identification
$$
E_1(K)=\{(t,w(t)):v(t)\geq1\}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 125 / 3 / d / Solution

For $y^2=x^3+ax+b$, put $t=-x/y$ and $w=-1/y$. Then $x=t/w$, $y=-1/w$, and the curve equation becomes
$$
w=t^3+atw^2+bw^3.
$$
There is a unique solution $w(t)\in t^3K[[t]]$, and points in the kernel of reduction correspond to parameters of positive valuation.


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 Incoming links: Formal coordinates on a short Weierstrass curve