= Solution
Write $A=k[x,y]$ and $\mathfrak m=(x,y)$. Since $f(0,0)\ne0$, the <principal open subscheme> $D(f)=\operatorname{Spec}A_f$ contains $\mathfrak m$, and
$$
V=D(f)\setminus\{\mathfrak m\}.
$$
Cover $V$ by the two affine opens $D(xf)$ and $D(yf)$, whose intersection is $D(xyf)$. The degree-zero part of the resulting <Čech cohomology> complex gives
$$
H^0(V,\mathcal O_V)=A_{xf}\cap A_{yf}=A_f
$$
inside the <fraction field> of $A$. The last equality follows because $A_f$ is a <unique factorization domain> and a rational function regular after localizing at both $x$ and $y$ has no possible prime factor left in its denominator.
The same affine cover is acyclic, so its degree-one Čech group computes <sheaf cohomology> and gives
$$
H^1(V,\mathcal O_V)
\cong A_{xyf}/(A_{xf}+A_{yf}).
$$
Before localizing at $f$, the quotient
$$
Q=A_{xy}/(A_x+A_y)
$$
has the $k$-basis
$$
\{x^{-a}y^{-b}:a,b\geq1\}.
$$
Writing $f=c+h$ with $c=f(0,0)\in k^\times$ and $h\in(x,y)$, multiplication by $h$ is locally nilpotent on $Q$: for each negative monomial, a sufficiently high power of $(x,y)$ moves every term into $A_x+A_y$. Hence $c+h$ acts invertibly on $Q$ by a finite geometric series on each element. Localizing at $f$ therefore leaves $Q$ unchanged, and
$$
H^1(V,\mathcal O_V)\cong Q_f\cong Q.
$$
The displayed infinite basis proves that this vector space is infinite-dimensional.
Solved by gpt-5.6-sol high.
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