EDIT: Maybe exact closed forms are too difficult to obtain. However, can we at least bound the size of the coefficients? This would be important in error and convergence analysis.
Puiseux series have, today I learned, a lovely connection with algebraic geometry. One can formally say that the field of Puiseux series over any algebraically closed, characteristic zero field is the algebraic closure of the field of Laurent series for that field - reference. I haven't quite digested the proofs of this fact, but I did find here an informal description of an algorithm to identify the Puiseux series.
This algorithm is nice, and I'll showcase it on a curve I worked on (yes, I know this one can be analytically solved... but Puiseux series can expand any curve with coefficients Laurent series in $x$, and the majority of these don't admit nice closed form solution):
$$y^2-y-\frac{1}{x}=0$$
Consider the equation as a polynomial in $y$, with coefficients Laurent series in $x$. If the coefficients of $y$ are labelled $A_0+A_1y+A_2y^2$, let $\alpha_{0,1,2}$ denote the orders of the $A_{0,1,2}$. Here, we have $\alpha_0=-1,\alpha_{1,2}=0$. Join the points $(k,\alpha_k)$ with lines to form a convex polygon - the Newton polygon - in the Cartesian plane; the line with the most negative slope has slope $\frac{\alpha_2-\alpha_0}{2-0}=\frac{1}{2}$, and we take the negative of this, $-\frac{1}{2}$, as our first exponent. The coefficient of this, $c$, will be a solution to $\sum_k a_kc^k=0$ where we sum over those $k$ which have $(k,\alpha_k)$ on the most negatively sloped line and $a_k$ is the coefficient in $A_k$ of least order. Here, that means $(-1)c^0+(1)c^2=0$ and we can take one branch of the solution as $c=1$. Thus, $x^{-1/2}$ is our first term.
Substitute $y\mapsto y+x^{-1/2}$ into the desired curve, to get the new equation: $$y^2+(2x^{-1/2}-1)y-x^{-1/2}=0$$We repeat the algorithm. $\alpha_0=-\frac{1}{2},\alpha_1=-\frac{1}{2},\alpha_2=0$ and the most negative slope is the line of slope zero connecting $(0,\alpha_0)$ and $(1,\alpha_1)$; thus our next exponent is $-0=0$. The coefficient $c$ must satisfy $(-1)c^0+(2)c^1=0,\,c=\frac{1}{2}$. So far, $x^{-1/2}+\frac{1}{2}$ is our solution.
Substitute $y\mapsto y+\frac{1}{2}$ in the above to obtain a third curve: $$y^2+2x^{-1/2}y-\frac{1}{4}=0$$And now $\alpha_0=0,\alpha_1=-\frac{1}{2},\alpha_2=0$ with most negative slope the line connecting $(0,\alpha_0)$ to $(1,\alpha_1)$, with slope $-\frac{1}{2}$. The next exponent is then $\frac{1}{2}$, and the coefficient $c$ must satisfy $-\frac{1}{4}+2c=0,\,c=\frac{1}{8}$
That gives $y\sim x^{-1/2}+\frac{1}{2}+\frac{1}{8}x^{1/2}$ as one branch of the curve. These coefficients are most efficiently found from noting: $$y=\frac{1\pm\sqrt{1+4x^{-1}}}{2}=\frac{1}{2}\pm x^{-1/2}\sqrt{\frac{x}{2}+1}$$And expanding in a Taylor series. As remarked earlier, this is not always possible. You could hand me as ghastly a curve as you liked, say a famous unsolvable-with-radicals quintic curve, and I would then be forced to plod through the Newton polygon algorithm to find the terms of the series. This is impractical for analysis: we generally like to know our Taylor/Laurent/Maclaurin series with coefficients in closed-form, for whatever purpose.
My question: Is there a way to, in general, identify a closed form for Puiseux coefficients? Since a Puiseux series $g$ is just some (complex) analytic $f$ with input $z^{1/n}$, so $g(z^n)$ is analytic, that suggests closed forms could be possible. This is reminiscent of using the implicit function theorem to identify local solutions to curves, but this has the same difficulty. See at the bottom of this post of mine: it took an hour of sneaky scribbling during class to find that $10$th order expansion to the implicit curve, and it was a relatively simple curve too... can we do better?