Example II.6.5.2 in Hartshorne (Part I)

As I said in the comment, the closed subset $Z(y)$ of zeros of $y$ is equal to $Y$ because $y=0$ implies that $z=0$. If you prefer, $\sqrt{yA}=(y,z)$.

So the support of $\mathrm{div}(y)$ is $Y$. To compute the multiplicity of this divisor, we consider the generic point $\eta$ of $Y$ and have to compute the valuation $v_{\eta}(y)$, where $v_{\eta}$ is the valuation on $K(X)$ (field of rational functions on $X$) defined by the discrete valuation ring $O_{X, \eta}$.

Let $\mathfrak p=(y, z)$ be the prime ideal of $A$ defining the point $\eta$. Then $\mathfrak p A_{\mathfrak p}$ is generated by $z$ because $y=z^2x^{-1}$. Therefore $v_{\eta}(y)=2v_{\eta}(z)=2$. So $\mathrm{div}(y)=2Y$.

Isn't $Y = \operatorname{Spec} k[x,y,z]/(xy-z^2,y,z) = \operatorname{Spec} k[x]$? So it is an integral closed subscheme of $X$ of codimension $1$.