Write $R=k[x_{1},\ldots,x_n]/(f_{1},\ldots,f_{c})$, $I=(f_{1},\ldots,f_{c})$, if $\Omega_{R/k}=0$, then $R$ must has dimension zero hence an Artinian ring which has finite many maximal ideals. For every maximal ideal $\mathfrak{m}$ we have a exact sequence $ \mathfrak{m}/\mathfrak{m}^{2}\to \Omega_{R/k}\otimes_{R} \kappa(\mathfrak{m})\to \Omega_{\kappa(\mathfrak{m})/k}\to 0. $ By assumption $\Omega_{R/k}=0$, it follows that $\kappa(\mathfrak{m})/k$ is a finite separble algebraic extension. So our exact sequence can become $ 0\to \mathfrak{m}/\mathfrak{m}^{2}\to \Omega_{R/k}\otimes_{R} \kappa(\mathfrak{m})\to \Omega_{\kappa(\mathfrak{m})/k}\to 0. $ It follows that $I_{\mathfrak{m}}=\mathfrak{m}_{\mathfrak{m}}$ because $\mathfrak{m}/\mathfrak{m}^2=0 $ and NAK, so $R=\prod R_{\mathfrak{m}}=\prod \kappa(\mathfrak{m})$.
If we assume that $R$ is a local ring, we can also directly deduce that $R$ is an Artinian local ring without using the hypothesis $\Omega_{R/k}=0$, anyway we prove the result.