Let $k$ such that $\operatorname{char}(k)=0$ and $C\subset \mathbb{P}^2$ be the smooth projective curve defined by $$X^4+Y^4+Z^4=0$$ I want to compute the divisor of $d\left(\frac{X}{Z}\right)$.
My approach is the following. First look at the points where $Z=1$. If we look at the affine plane $\mathbb{A}^2\cong \{Z=1\}\subset \mathbb{P}^2$, these are points of the form $(a,b)$. The maximal ideal corresponding to these points is $(X-a,Y-b)$. I think a uniformizer at any of these points can be $(X-a)$. We also have that since $Z=1$, $d\left(\frac{X}{Z}\right)=dX$. Thus, for these points we can say that $$\operatorname{ord}_P\left(d\left(\frac{X}{Z}\right)\right)=\operatorname{ord}_P\left(dX\right)=\operatorname{ord}_P\left(d(X-a)\right):=\operatorname{ord}_P\left(\frac{d(X-a)}{d(X-a)}\right)=0.$$ Now we go to the points $Z= 0$. These will be points of the form $(a:1:0)$. If we look at the affine plane $\mathbb{A}^2\cong \{Y=1\}\subset \mathbb{P}^2$, these are points of the form $(a,1)$. The maximal ideal corresponding to these points is $(X-a,Y-1)$. I think I can take $\frac{1}{Y}$ as a uniformizer. Looking at the equation $X^4+1=0$, I know that this can only happen when $X=\pm\sqrt{-1}$, so the only points here in this case will be $(\sqrt{-1}:1:0)$ and $(-\sqrt{-1}:1:0)$. However, I am not sure where to go from here. Could someoene help me?