Consider the exact sequence $0 \to S(-\mathrm{deg}\; f) \to S \to S/(f) \to 0$ (where the first map is multiplication by $f$) and take its long exact sequence of $\mathrm{Ext}$ groups. Since both $S$ and $S(-\mathrm{deg}\; f)$ are free $S$-modules, their higher $\mathrm{Ext}$ groups vanish, and you get $\mathrm{Ext}^m(S/(f), S) = 0$ for all $m \geq 1$2$. In addition, it is clear that$\mathrm{Ext}^0(S/(f), S) = \mathrm{Hom}(S/(f), S) = 0$. 1 Consider the exact sequence$0 \to S(-\mathrm{deg}\; f) \to S \to S/(f) \to 0$(where the first map is multiplication by$f$) and take its long exact sequence of$\mathrm{Ext}$groups. Since both$S$and$S(-\mathrm{deg}\; f)$are free$S$-modules, their higher$\mathrm{Ext}$groups vanish, and you get$\mathrm{Ext}^m(S/(f), S) = 0$for all$m \geq 1$. In addition, it is clear that$\mathrm{Ext}^0(S/(f), S) = \mathrm{Hom}(S/(f), S) = 0\$.