Confusion on the divisor of the $1$-form $(dx)$ on $y^3-x^5+1$

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Consider the curve $y^3-x^5+1$ and its compactification $C$ in $\mathbb{P}^2$ given by $z^2y^3-x^5+z^5$ and one additional point $p$ (not its projective closure). Following Geometry of Algebraic Curves, I want to

a) Find $(dx)$,

b) Use this to describe $H^0(K_C)$, for $K_C$ the canonical line bundle (dual of the tangent bundle).


a) I know that $ord_p(dx):=ord_p(f)$ where $dx=fdj$ in some locally holomorphic coordinate $j$. We can see that $(x)$ contains points at $[0:w:-1]$ and the point at infinity, for $w^5=1$. However, these are zeroes not poles, so I believe that $x$ is everywhere holomorphic on $C$ (am I wrong?) and thus that $x$ is always a locally holomorphic coordinate and thus $ord_p(dx)=0$ uniformly. Does this mean $(dx)=0$?

b) If so, how can this be used to compute $H^0(K_C)$? I know that it is the space of holomorphic $1$-forms, and I further know that all holomorphic $1$-forms are linearly equivalent ($fdj/hdj=f/h$).

My reservations: My computation of $(dx)$ does not seem dependent on the curve. I'm not sure how to describe $H^0(K_C)$ in general.