I was trying to do this exercise, and happened to find a solution, but something bothers me.
Let $K$ be a field and let $C$ be the smooth projective curve given by the equation $y^2 = xz$ in $\mathbb{P}^2$.
(a) Compute the divisor of the function $y/z$.
This yields $\big(\frac yz\big)=[0:0:1]-[1:0:0]$ .
(b) Compute the divisor of the function $x/z$.
This yields $\big(\frac xz\big)=2[0:0:1]-2[1:0:0]$
(c) Find a function whose divisor is $[1:1:1]−[0:0:1]$.
For this, I was following Example 11.4 of this script, and proceeded as follows:
I want $D=[1:1:1]−[0:0:1]$. The line through the points $(0,0)$ and $(1,1)$ is given by $y-x=0$. Homogenizing and looking for zeros and poles gives: $\big(\frac{y-x}z\big)=[1:1:1]+[0:0:1]-2[1:0:0]$. Now, the vertical line through $(1,1)$ (why this point?) is $x-1=0$. Homogenizing and looking for zeros/poles, yields $\big(\frac{x-z}z\big)=2[1:1:1]−2[1:0:0]$
Then notice that $D=[1:1:1]−[0:0:1]=2[1:1:1]−2[1:0:0]-[1:1:1]-[0:0:1]+2[1:0:0]=\big(\frac{x-z}z\big)-\big(\frac{y-x}z\big)=\big(\frac{x-z}{y-x}\big)$
So this the function I was looking for (I also checked after, that this function gives the desired divisor). But in the example I referred to, one has the divisor of the line through the two points we calculated first, minus the divisor of the vertical line, while I did the opposite (just because I noticed that it would give me what I was looking for).
Is it a "nice" recipe to solve this kind of exercise? Are part (a) and (b) relevant for part (c), in the sense that those parts could have helped me better for part (c)?
In the beginning of the example it says "The proof of the theorem gives an algorithm for finding a function with a given divisor", but I didn't spend too much on understanding what was it referring about (I still had a look at the proof in question), instead tried following directly the procedure of the example.
For smooth conics, the recipe is pretty quick. Let $p,q\in C$ be distinct points, and let $r\in C$ be a point distinct from $p,q$. Let $L_1=V(h_1)$ be the line through $p$ and $r$, and let $L_2=V(h_2)$ be the line through $q$ and $r$. Then $(\frac{h_1}{h_2})=p-q$, as $h_1$ has a zero of order $1$ at both $p$ and $r$ while $h_2$ has a zero of order $1$ at both $q$ and $r$. This shows that on a smooth conic, any divisor of the form $\sum_{i=1}^n p_i - \sum_{i=1}^n q_i$ is the divisor of a rational function by applying this process to each of the pairs $p_i-q_i$.
Parts (a) and (b) are useful because they give you a little practice. One thing you could learn from there which would be helpful for the second part is noting that the zeroes of a linear function are either two points or one point with multiplicity two, depending on whether the intersections are transverse or a tangent line ($V(y)$ intersects the conic at $[0:0:1]$ and $[1:0:0]$, $V(x)$ intersects the conic at $[0:0:1]$ twice).
The example from the script has a fairly different character - elliptic curves have a rich structure where you can play around but it's still pretty bounded in terms of difficulty - not every degree-zero divisor is the divisor of a rational function, but you can tell easily by adding everything up in the group law. Smooth conics are verrrrry simple - every degree-zero divisor is the divisor of a rational function, and smooth higher-degree curves are a bit more interesting because not every degree-zero divisor is the divisor of a rational function, and there's not the easy "add all the points" strategy to tell.