I am confused about the Definition 4.1.1 and Exercise 4.1.A in Vakil's notes on Algebraic Geometry.
4.1.1. Definition. Define $\mathcal{O}_{\mathrm{Spec}A}(D(f))$ to be the localization of $A$ at the multiplicative set of all functions that do not vanish outside of $V(f)$ (i.e., those $g\in A$ such that $V(g)\subset V(f)$, or equivalently $D(f)\subset D(g)$). This depends only on $D(f)$, and not on $f$ itself.
4.1.A. Exercise. Show that the natural map $A_f\rightarrow \mathcal{O}_{\mathrm{Spec}A}(D(f))$ is an isomorphism.
If we let $A=k[x,y]$, $f=xy$, and $g=x^2y$, then $g$ is not in the multiplicative set $\{1,f,f^2,\dots\}$, so $1/g$ is not in $A_f$. But by Definition 4.1.1, it is in $\mathcal{O}_{\mathrm{Spec}A}(D(f))$ since $D(f)\subset D(g)$, or $V(g)\subset V(f)$.
Since the isomorphism seems not to work for this example, I thought maybe my idea of $A_f$ is not correct. But I found in Chapter 1, after 1.3.3,
There are two particularly important flavors of multiplicative subsets. The first is $\{1,f,f^2,\dots\}$, where $f\in A$. This localization is denoted $A_f$.
By this, I think $1/g$ is not in $A_f$. What am I missing? Any help would be appreciated!
$\frac{1}{g}=\frac{1}{x^2y}=\frac{y}{x^2y^2}=\frac{y}{f^2}\in A_f$