Hyperplane divisors: working on an example. I'm working Problem H of section I of chapter V of Miranda's book. The problem says

Let $X$ be the smooth projective plane curve defined by $y^2z=x^3-xz^2$. Compute the intersection divisors of the lines $x=0$, $y=0$ and $z=0$ with $X$.

Roughly speaking, take for example the intersection with $y=0$:
$$
\begin{cases}
y^2z=x^3-xz^2 \\ x=0
\end{cases} \quad \Longrightarrow \quad x(x^2-z^2)=0 \quad \Longrightarrow x=0, x=\pm z
$$
so I get the points $p_1=[0:0:1]$, $p_2=[1:0:1]$ and $p_3=[1:0:-1]$. So I'm expect to get
$$div(y)=p_1+p_2+p_3$$ 
and it works in this case: I take for $p_1$ the homogeneous polynomial of degree 1 $H=z$, so $H(p_1)\ne 0$ and I can compute
$$div(y)(p_1)=ord_{p_1}(x/z)=ord_{p_1}(x)-ord_{p_1}(z)=1-0=1.$$
I repeat the computation also for $p_2$ and $p_3$ and it's all ok.
The problem come up with $x=0$:
$$\begin{cases} y^2z=x^3-xz^2 \\ x=0 \end{cases} \quad \Longrightarrow y^2z=0
$$ 
so I get $p_1=[0:0:1]$ (counted two times, like in the affine real picture) and $p_2=[0:1:0]$. So I expect that my divisor will be in the form
$$div(x)=2p_1+p_2.$$
Like before, I take $H=z$ so $H(p_1)\ne 0$ and compute using the definition:
$$div(x)(p_1)=ord_{p_1}(x/z)=1.$$
Where am I wrong? This it makes me going crazy.

 A: I just worked through this problem (I believe) and for posterity's sake I figured I'd post some more information here. Let's write $F(x,y,z)=y^2z-x^3+xz^2$ so that $X=\mathcal{Z}(F).$ Then 
$$ \frac{\partial F}{\partial x}(x,y,z)=-3x^2+z^2,\:\:\:\:\:\frac{\partial F}{\partial y}(x,y,z)=2yz,\:\:\:\:\:\frac{\partial F}{\partial z}(x,y,z)=y^2+2xz.$$
Let us study first the case of $y=0$. It is easy to see that the resulting equation is $-x^3+xz^2=0$ whence $x=0$ or $x^2=z^2$, so $x=\pm z$. Then, the solutions are 
$$p_1=[0:0:1],\:\:\:\:\: p_2=[1:0:1],\:\:\:\:\:\: p_3=[1:0:-1].$$ 
We begin with  $p_1$. We can see that our complex parameters are $(\frac{x}{z},\frac{y}{z})\in \mathbb{C}^2$. Further, $\partial_x(F)(p_1)=1$ and $\partial_y(F)(p_1)=0$ so that $\frac{y}{z}$ is the local coordinate on $X$ at $p_1$ by the Implicit Function Theorem. As for calculating the hyperplane divisor, study the meromorphic function $\frac{y}{z}$. Applying the coordinate transformation, $y\mapsto \frac{y}{z}$ and $z\mapsto 1$, so in local coordinates on $X$, we have that our meromorphic function $\frac{y}{z}$ is $\frac{y}{z}$ in the local parameter $\frac{y}{z}$. Hence, the order here is $1$. We can apply the  same reasoning for $p_2, p_3$ and we get that 
$$ \operatorname{div}(y=0)=1\cdot p_1+1\cdot p_2+1\cdot p_3.$$
Onto  the case of $x=0$. If we have $x=0$, then we are left with $y^2z=0$ so that the points of intersection are $[0:0:1]=p_1$ and $[0:1:0]=p_4$. Let's examine $p_1$. $\partial_x(F)(p_1)=1$ and $\partial_y(F)(p_4)=0$. So, again $\frac{y}{z}$ is the local parameter. Study the meromorphic function on $\mathbb{P}^2$ given by $\frac{x}{z}$.
$$ \frac{x}{z}=\frac{x}{\frac{x^3-xz^2}{y^2}}=\frac{y^2x}{x^3-xz^2}=\frac{y^2}{z^2}\cdot \frac{1}{\frac{x^2}{z^2}-1}.$$
By the Implicit Function Theorem, for a holomorphic function vanishing at $0$, we can write $\frac{x}{z}=h(\frac{y}{z})$, so 
$$ \frac{x}{z}=\frac{y^2}{z^2}\cdot \frac{1}{h(\frac{y}{z})^2-1}$$
has a zero of order $2$. Another routine calculation shows that at $p_4$ $\frac{x}{y}$ has a simple zero. So, 
$$ \operatorname{div}(x=0)=2\cdot p_1+1\cdot p_4.$$
Last, we consider the  case $z=0$. Then we have an equation $x^3=0$, so $x=0$. Thus, the only point of intersection of $X$ with the hyperplane $z=0$ is $p_4=[0:1:0]$. $\partial_x(F)(p_4)=0$ and $\partial_z(F)(p_4)=1$ so that the local parameter is $\frac{x}{y}$. Write $g(\frac{x}{y})=\frac{z}{y}$, for $g$ holomorphic and vanishing at $0$. Now, consider the meromorphic function $\frac{z}{y}$.
$$ \frac{z}{y}=\frac{x^3-xz^2}{y^3}=\frac{x^3}{y^3}-\frac{xz^2}{y^3}=\left(\frac{x}{y}\right)^3-\frac{x}{y}\cdot \left[g\left(\frac{x}{y}\right)\right]^2.$$
By the implicit function theorem, we know (writing $\widetilde{x}=\frac{x}{y}$ and $\widetilde{z}=\frac{z}{y}$) that 
$$ \frac{\partial g}{\partial \widetilde{x}}(0)=-\frac{\partial_{\widetilde{x}}(f)}{\partial_{\widetilde{z}}(f)}(0,0)=0.$$
So, $[g(\widetilde{x})]^2$ vanishes at least to order $4$. Hence,
$\operatorname{ord}_0(\frac{z}{y})=3.$ Thus,
$$ \operatorname{div}(z=0)=3\cdot p_4.$$ 
Notice, of course, that this exemplifies the result that the intersection divisor of a degree $d$ projective plane curve with a line in $\mathbb{P}^2$ has  intersection divisor of degree $3$. Lastly, note that this approach uses the fact that $X$  is cut out by $F(x,y,z)$. 
