I want to compute the degree of the twisted cubic curve $$X = \{xw = yz, xz=y^2, yw = z^2\} \subset \mathbb{C}\mathbb{P}^3 \ni [x:y:z:w].$$ I already know that $\deg(X) = 3$, however, I want to verify this claim by computing the degree of the intersection divisor $\operatorname{div}(x)$.
To this end, let $L: x= 0$. Then,
$$ X \cap L = \{yz = 0, y^2=0, yw = z^2\} $$ contains only the point $p = [0:0:0:1]$, since $y^2 = 0$ implies $y = 0$ and hence, $z = 0$. If it is true that there is a unique intersection point, then, by definition, the degree of $X$ is the order of the meromorphic function $\tfrac{x}{w}$ at $p \in X$. Why is this equal to 3? The order is $2$ around $y^2 = 0$ and likewise, around $z^2 = 0$. I am not sure how one comes up with 3.
By definition $$ \deg(X) = \ell(X \cap L), $$ the length of the finite scheme $X \cap L$.
To identify this scheme in $L \cong \mathbb{P}^2_{(y:z:w)}$ note that it lies in the affine chart $w \ne 0$, so we may take $w = 1$ and use $y$ and $z$ as coordinates. Then the equations become $$ yz = 0, \quad y^2 = 0, \quad y = z^2. $$ Using $y = z^2$ to eleminate $y$ we obtain $z^3 = z^4 = 0$ from the other two equations, hence $$ X \cap L \cong \mathrm{Spec}(\mathbb{C}[z]/z^3), $$ hence its length is 3.