Jacobian ideal of $f$ and normality 
Let $k$ be an algebraically closed field of characteristic $0$. Suppose $f\in k[X_1,\ldots,X_n]$ is such that $f\in \mathcal{J}(f)$, where $\mathcal{J}(f)$ is the Jacobian ideal of $f$ (i.e. ideal generated by the partial derivatives of $f$). Then is it always true that $k[X_1,\ldots,X_n]/\langle f+1 \rangle$ is a normal affine algebra? (Note: I am looking at the hypersurface $f+1$.)

A special case: if $f$ has isolated singularity, then by Saito's result $f$ is quasi-homogeneous and hence, in fact $k[X_1,\ldots,X_n]/\langle f+1 \rangle$ will be smooth.
Ref: Kyoji Saito, Quasihomogene isolierte Singularitäten von Hyperflächen, Invent. Math. 14 (1971), 123–142. MR0294699
 A: The Jacobian criterion for non-singularity and Serre's criterion for nomality will be handy in this case.

Jacobian criterion.
For $A = k[x_1,\dots, x_n]$ and $R = A/ (f)$ for some non zero $f \in A$, the nonsingular locus is defined by the ideal $\mathrm{Jac}(f)$, the ideal generated by all partials of $f$, extended to $R$.



Serre's criterion.
Such $R$ above is normal iff $R$ satisfies Serre's condition $(R_1)$, i.e., $R_p$ is regular for all prime ideals of codimension $1$.
(I somehow recall that to say a ring is normal, the ring needs to be a domain, i.e., $f$ is a prime.)

To show that $A' = A/(f+1)$ is normal given that $f \in \mathrm{Jac}(f)$ it suffices to show that codimension of $\mathrm{Jac}(f+1)A'$ is at least $2$. Indeed the condition $f \in \mathrm{Jac}(f)$ implies that $\mathrm{Jac}(f)A' = A'$ as $\mathrm{Jac}(f) = \mathrm{Jac}(f+1)$. Hence $A'$ is normal.
A: Your hypersurface with equation $f(X_1,...,X_n)+1=0$ is certainly normal because it is nonsingular.
But why is it nonsingular?
Because its singularities are given by the vanishing locus of the $n+1$ equations
$$f+1=0,\;\frac {\partial f}{\partial X_1}=0\;,...,\;\frac {\partial f}{\partial X_n}=0$$
And then the last  $n$ equations force $f=0$ because of your hypothesis that $f$ is in the ideal generated by the Jacobian ideal of $f$.
So any singular point $s$ of your hypersurface $f+1=0$ would have to satisfy $f(s)=0$ and $f(s)+1=0$: thus no such $s$ exists and we have the claimed nonsingularity.
