Let X be a smooth curve in the projective plane such that for $n \in \mathbb{Z} \geq 2$ we have that $X=Z(-x^n_1+x_0^{n-1}x_2-x^n_2)$. We assume that $n(n-1)$ is in $k^{\times}$. We know that $X$ is smooth and irreducible. We also let $U= Z(f)$ where $f=-y^n=x^{n-1}-1$ Now we have found that $$\omega_0= \frac{dx}{ny^{n-1}}=\frac{dy}{(n-1)x^{n-2}}$$ has order $0$ for all points of $U$. And we conclude that $\omega_0$ is a basis for $\Omega^1(U)$ as a free $\mathcal{O}(U)-$module.
We have also calculated that $Q=[1:0:0]$ a unique point at infinity, and that the uniformizer of $U$ is $x_{0,1}$.
We then consider the chain $(x,y) \rightarrow [x:y:1] \rightarrow (y/x, 1/x)$ where $(x,y)=U_2$ and $(y/x, 1/x)=U_0$. We then written $x,y, \omega_0$ usnign the $U_0$ chart, to find $\nu_Q(x)=-n$, $\nu_Q(y)=(1-n)$, and $\nu_Q(\omega_0)=(n-3)n$.
We now want to find a basis for $\Omega^1(X)$ for $n=2,3,4$. I believe that we have $\omega_0$ as a basis for $\Omega^1(X)$ when $n=3$, as the valuation at $Q$ is zero in that case. And I also believe that $Q \in U_0$ so it is also taken care of in this case.
I am however lost on how to find a basis $n=2,4$. I think that the correct route is to find an $\omega_0$ such that $\nu_Q(\omega_0)=0$. But I don't know now how to do this. I am also wondering in general how to think about a basis for $\Omega^1(X)$. I think another route is to find an open cover of $Q$ such that on this open cover intersected with the $U$ defined above the rational one forms agree. But I am also uncertain how to calculate this explicitly, and this argument is dependent on $Q$ being only point from $X$ not in $U$. All the best.
Thanks