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Put $A=k[X]$. There is an explicit isomorphism from the principal open subset $U_f$ to the affine algebraic setIts inverse is projection, and both maps are regular maps. The coordinate ring isby the universal property of localization. Since regular functions on an affine variety are its coordinate ring, this proves the natural $k$-algebra isomorphism

Now $W=D(x)\cup D(y)$. A regular function on $W$ determines one element of the function field $k(x,y)$, and its restrictions belong to $k[x,y{]}_x$ and $k[x,y{]}_y$. Conversely, elements of their intersection glue because they agree on $D(xy)$. ThusIf a reduced fraction has denominator dividing a power of $x$ and also a power of $y$, unique factorization makes its denominator a unit. Therefore puncturing the plane does not change its ring of global regular functions:If $W$ were an affine variety, its inclusion $j:W\hookrightarrow {\mathbb{A}}^2$ would correspond to the identity isomorphism $k[x,y]\to k[W]$. The equivalence between affine varieties and their coordinate rings would make $j$ an isomorphism, contradicting the missing origin. Hence $W$ is not affine. This illustrates regular functions on the punctured affine plane: the coordinate ring alone does not recover a nonaffine variety.