Uniformiser in book by Miranda I am doing Exercise V.2D in Miranda's book Algebraic Curves and Riemann Surfaces.

Let $X$ be the projective plane curve defined by the equation $y^2 z=x^3-xz^2$. Let $p_0=[0:1:0]$, $p_1=[0:0:1]$, $p_2=[1:0:1]$ and $p_3=[-1:0:1]$. Show that $p_1+p_2+p_3\sim 3p_0$.

Taking $f=Y/Z$, the points where it has zeros and poles are precisely the $p_i$ above.
When I want to calculate the order of vanishing at $[0:0:1]$, I go to the affine chart $Z\neq 0$, then the point becomes $(0,0)$ and I have to calculate $ord_{(0,0)}(y)$. I believe this is one, because there is no power in the $y$. However, I know from the theory of uniformisers in algebraic geometry that this is not necessarily the case. For example, when I do the same trick for $
[0:1:0]$  then the order of vanishing should be -3 but is $ord_{(0,0)}(1/z)=-1$ when I reason like this.
I can not find anywhere in Miranda's book the theorem that says that a function that vanishes at a point but is not the tangent line is a uniformiser. Could someone help me out ??
Also, in Miranda's book it seems that the order of vanishing is defined by the lowest non-zero term in the Taylor expansion, but isn't this $y$ ?
 A: Miranda's book has more of a complex geometry flavor, so he defines order of vanishing using holomorphic charts and local coordinates. (It's really not that different from using uniformizers and local rings, though.) The analogue of the result you're looking for is the passage preceding Theorem 2.3 of Chapter I (p. 11), which says more or less the following.
Proposition. Let $C$ be a nonsingular affine plane curve given by the equation $f(x,y)=0$, where $f \in \mathbb{C}[x,y]$ is irreducible. Given $P \in C$, if $\frac{\partial f}{\partial y}(P) \neq 0$ then $x$ is a local coordinate on some neighborhood $U$ of $P$. In other words $(U, \pi_x)$ is a holomorphic chart containing $P$, where $\pi_x : (x,y) \mapsto x$.
(And similarly if $\frac{\partial f}{\partial x}(P) \neq 0$, then $y$ is a local coordinate near $P$.) The proof is basically the Implicit Function Theorem (Theorem I.2.1). Since the tangent line at the point $P = (a,b)$ is $\frac{\partial f}{\partial x}(P) (x-a) + \frac{\partial f}{\partial y}(P) (y-b)$, this agrees with the criterion you stated in terms of tangent lines.
Okay, so let's apply the proposition to your example. Let $E$ be the projective curve given by $Y^2 Z = X^3 - X Z^2$. (Forgive me for changing the notation, but I prefer to use lower case letters for affine coordinates.) In the affine open where $Z \neq 0$, $E$ is given by $y^2 = x^3 - x$, where $x = X/Z$ and $y = Y/Z$, so let $f = y^2 - (x^3 - x)$. Here $p_2 = (1,0)$, and since $f_x(p_2) = -(3(1)^2 - 1) \neq 0$, then $y$ is a local coordinate near $p_2$, so $\DeclareMathOperator{\ord}{ord} \ord_{p_2}(y) = 1$. (As you said, its Taylor series is simply $y$!)
To compute the order of vanshing at $p_0 = [0:1:0]$, we move to the affine open where $Y \neq 0$. Letting $w = X/Y$ and $z = Z/Y$, then $E$ is given by $z = w^3 - w z^2$, and $p_0 = (0,0)$. From this equation we can already see that $z$ can't be a uniformizer at $p_0$ since $z = w^3 - w z^2$ and both terms on the righthand side vanish to at least order $3$. Letting $h = z - (w^3 - wz^2)$, then
$$
h_z(p_0) = 1 - (-2wz)\big{|}_{(w,z) = (0,0)} = 1 \neq 0
$$
so $w$ is a local coordinate near $p_0$. To compute $\ord_{p_0}(z)$ we need to express $z$ as a Taylor series in $w$. One way to do this is by repeatedly substituting $z = w^3 - w z^2$ into the righthand side of the equation:
\begin{align*}
z &= w^3 - w z^2 = w^3 - w (w^3 - w z^2)^2 = w^3 - w^7 + 2 w^5 z^2 - w^3 z^4\\
&= w^3 - w^7 + 2 w^{11} - 5 w^{15} + \cdots
\end{align*}
If you don't like that approach, the Implicit Function Theorem implies that $z = g(w)$ for some function $g$ holomorphic near $w=0$. Substituting this into the equation for the curve yields
$$
g(w) = w^3 - w \, g(w)^2
$$
and we can compute as many Taylor coefficients as we like by repeatedly differentiating and substituting $w = 0$.
In any case, we find that
$$
\ord_{p_0}(y) = \ord_{p_0}(1/z) = -\ord_{p_0}(z) = -3 \, .
$$
