example of computing ramification index

Mohan essentially gives the answer in his comment above. But I figured I would write it up as an answer to my question.

Note that the fibre of $\phi$ at the point $[1,0]$ is precisely the point $[1,0]$. Thus restricting $\phi$ to the affine chart given by $Y=1$ gives a map $$ \phi: \mathbb{A}^1\to \mathbb{A}^1 $$

and in this case $\phi$ is precisely the polynomial $x^3(x-1)^2$. Now in the field $K(x)$, a uniformizer for $0\in \mathbb{A}^1$ is $x$ and a uniformizer for $1\in \mathbb{A}^1$ is $x-1$. Thus, $$ ord_{[0,1]}(\phi) = ord_0(x^3(x-1)^2) = 3 $$ and similarly for the order of $\phi$ at $[1,1]$.