Dear Charles, here is a more general statement:

Let $R$ be a reduced Noetherian ring of dimension 1. Then the set of singular points of $Spec(R)$ is finite.

Proof: Let $V$ be the set of singular point, i.e $V=\{p \| R_p \text{is singular} \}$. Then V does not contain any minimal prime of $R$, since $R$ localize at them are fields (as $R$ is reduced). So $\dim V=0$, and $V$ is closed, therefor $V=Spec(R/J)$ with $dim R/J=0$. But it is well known that an Artinian ring has only finitely many primes.

In your situation, let $R=k[x,y]/(f)$. Since $f$ is irreducible, $R$ is a domain and therefore reduced (reducedness = $f$ has no repeated factors which is weaker then irreducible). Any point that is $0$ on $f,f_x,f_y$ will give a maximal singular point in $Spec(R)$. So $S(f)$ is finite.

Dear Charles, here is a more general statement:

Let $R$ be a reduced affine algebra over $k$ of dimension 1. Then the set of singular points of $Spec(R)$ is finite.

Proof: Let $V$ be the set of singular point, i.e $V=\{p \| R_p \text{is singular} \}$. Then V does not contain any minimal prime of $R$, since $R$ localize at them are fields (as $R$ is reduced). So $\dim V=0$, and $V$ is closed, therefor $V=Spec(R/J)$ with $dim R/J=0$. But it is well known that an Artinian ring has only finitely many primes.

In your situation, let $R=k[x,y]/(f)$. Since $f$ is irreducible, $R$ is a domain and therefore reduced (reducedness = $f$ has no repeated factors which is weaker then irreducible). Any point that is $0$ on $f,f_x,f_y$ will give a maximal singular point in $Spec(R)$. So $S(f)$ is finite.