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I was thinking by looking at this as a polynomial in x over $\mathbb{C}[y]$. And then trying to use Gauss Lemma, Einstenstin criteria etc but that didn't work.

NumDio
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    So in general this problem is really hard, but has been discussed on MO here: https://mathoverflow.net/questions/105304/criteria-for-irreducibility-of-polynomial/105323#105323

    Although in your case I believe it's pretty easy to check it explicitly. I'll update if I find a proof.

     – A. Thomas Yerger Oct 02 '21 at 03:50
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    See here and here – leoli1 Oct 02 '21 at 08:03

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This polynomial is irreducible if and only if the ideal it defines is prime, so let's check that the resulting quotient ring is an integral domain. This is easy because in our setup, we can write $\mathbb{C}[x,y]/I = \mathbb{C}[x,y]/(x^{13} - y^{11}) \simeq \mathbb{C}[x,x^{13/11}]$ consisting of "polynomials" in $x$ that are allowed to have powers of $13/11$ as their exponents in addition to whole numbers. This representation makes it clear that there are no zero divisors.

To understand what's going on with general expressions of the form $y^n = p(x)$, there is a relatively simple machinery connecting this algebraic curve/polynomial ring to the ring of Puiseux Series. Roughly speaking, the idea is that you resolve any singularities, such as at $0$ in your curve, and write down a function on the resolved curve. Then you use the equation of the curve to write this as a function on the original curve, which will involve polynomials/power series with these fractional powers.

A. Thomas Yerger
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    The ring $\mathbb{C}[x,x^{13/11}]$ is not the same as $\mathbb{C}[x,x^{2/11}]$; $x^{2/11}$ is contained in the latter, but not in the former. – Jun Koizumi Oct 02 '21 at 11:20
  • Oh you're right. Warning kids, don't drink and derive. Anyway, logic still works out just fine for the integral domain. – A. Thomas Yerger Oct 02 '21 at 16:54