Finding the principal divisors 
Find the principal divisors of $g=x$ in $K(C)$

Given an elliptic curve $C:y^2=x^3+1 \pmod {13}$. How can I find the principal divisor of the function $g=x$. Upon homogenizing I get $g=xz$, $y^2z=x^3+z^3$ now $(0:-1:1),(0:1:1),(0:1:0)$ are the special points, (Does Bézout theorem state that, there are $3$ such points ?) At first sight, $g$ doesn't seem to have any poles, but it should and I guess the coefficient of $(0:1:0)$ is $-2$. I wanted to solve this in the same manner as here (Example $11.1$), but then $x=\frac{y^2}{x^2+\frac1x}$. It doesn't go well. 
 A: Here are two ways to find the answer for the divisor of $x$, one more theoretical, and the other more computational.
Theoretical:
The map $x: E \to \mathbb{P}^1$, $(x,y)\mapsto x$ has degree $2$, because for each fixed value $x= x_0$ in the field $k$, the equation $y^2 = x_0^3 - 1$, i.e., $y^2 - x_0^3 + 1 = 0$ has $2$ solutions (for $y$), counting multiplicity.  This shows that $x$ is generically $2$ to $1$.
We see that, in the affine open set $Z \neq 0$ where $E$ is given by $y^2 = x^3 + 1$, $x$ has no poles.  Thus all poles must occur on the line at infinity $Z = 0$.  Since $E$ only meets the line at infinity at $[0:1:0]$ (to see this, substitute $Z = 0$ into $Y^2Z = X^3 + Z^3$), then $x$ must have a pole at $[0:1:0]$.  Since $x$ has degree $2$, and this is the only pole, then it must be of order $2$.
Computational: I always prefer to see with my own eyes that something is a pole.  As mentioned in the comments, since $x = X/Z$, we can consider $x$ as a map to $\mathbb{P}^1$ given by
\begin{align*}
x: E &\to \mathbb{P}^1\\
[X:Y:Z] &\mapsto [X:Z]
\end{align*}
The poles of $x$ are just the points on $E$ that map to $[1:0] = \infty$.  If $x([X:Y:Z]) = [X:Z] = [1:0]$, then $Z = 0$, and substituting this into the equation $Y^2Z = X^3 + Z^3$ yields $X^3 = 0$, so $X = 0$.  Thus the point $[0:1:0]$ maps to $[1:0]$.  This is not apparent from our expression for $x$; it looks like $[0:1:0]$ maps to $[0:0]$, which is not a point in projective space.  But the expression above is only valid on the set where $X \neq 0$ or $Z \neq 0$.  To find the expression instead where $Y \neq 0$, we note that
$$
[X:Z] = [X^3:X^2Z] = [Y^2 Z - Z^3: X^2Z] = [Y^2 - Z^2:X^2]
$$
since $Y^2Z = X^3 + Z^3$.  From this expression, we see that $[0:1:0]\mapsto [1:0]$.
To find the order of this pole as in the example you linked, we need to look in local coordinates around $[0:1:0]$ on the affine open $Y \neq 0$.  Let $u = X/Y, v = Z/Y$ be coordinates on this set.  Then $E$ is given by
\begin{align*}
v = \frac{Z}{Y} = \frac{Y^2 Z}{Y^3} = \frac{X^3 + Z^3}{Y^3} = \frac{X^3}{Y^3} + \frac{Z^3}{Y^3} = u^3 + v^3
\end{align*}
so $E: v = u^3 + v^3$ and the point $[0:1:0]$ corresponds to the point $(u,v) = (0,0)$.  Also, note that $x = X/Z = \frac{X/Y}{Z/Y} = u/v$.  One can see that $u$ is a uniformizer at $(u,v) = (0,0)$, and since
$$
v = u^3 + v^3 \implies u^3 = v - v^3 = v(1 - v^2) \implies v = \frac{u^3}{1 - v^2}
$$
then $\text{ord}_{(0,0)}(v) = 3$.  (This is just like the example you linked.)  Then
$$
\text{ord}_{(0,0)}(x) = \text{ord}_{(0,0)}(u/v) = \text{ord}_{(0,0)}(u) - \text{ord}_{(0,0)}(v) = 1 - 3 = -2
$$
so, as before, we see that $x$ has a double pole at $[0:1:0]$.
