= Solution
Let $U=\mathbb P^2\setminus C$. The <localization sequence for the divisor class group> gives
$$
\mathbb Z[C]\longrightarrow\operatorname{Cl}(\mathbb P^2)\longrightarrow\operatorname{Cl}(U)\longrightarrow0.
$$
Every <prime Weil divisor> of $U$ extends by closure to one of $\mathbb P^2$, and the only removed <prime Weil divisor> is $C$. Rational functions have the same function field on the two spaces, so the kernel on <divisor class groups> consists exactly of multiples of $[C]$.
The <divisor class group> of $\mathbb P^2$ is $\mathbb Z$, generated by the class $[H]$ of a line. To see the degree identification, if a plane curve $D$ has degree $e$ and homogeneous equation $F_D$, then $F_D/\ell^e$, for a line equation $\ell$, is a rational function with <principal Weil divisor> $D-eH$. Degrees of <principal Weil divisors> are zero, so $[H]$ has infinite order. In particular, $[C]=d[H]$. The localization sequence therefore yields
$$
\boxed{\operatorname{Cl}(\mathbb P^2\setminus C)\simeq\mathbb Z/d\mathbb Z.}
$$
This is the <divisor class group of a plane-curve complement>. It includes $d=1$, when the group is zero, and does not require the removed curve to be nonsingular.
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