Does $\det \mathcal{H}_f = (n-1)(-1)^{2n+1} \left[ f(x_1, \dots, x_n) \right]^{n-2}$ for $f(x_1, \dots, x_n) = \prod_{j=1}^n x_j$? Let function $f : \mathbb{R}^n \to \mathbb{R}$ be defined by $$f(x_1, \dots, x_n) := \prod_{j=1}^n x_j$$ Like any $C^2$ function, we can compute its Hessian $\mathcal{H}_f$, which will be a $\mathbb{R}^n \to \mathbb{R}^{n \times n}$ function. $\mathcal{H}_f$ should also have a determinant, denoted by $\det \mathcal{H}_f$.
I have computed $\det H_f$ for $n\in\{2,\dots, 8 \}$ and I have come up with a guess that is consistent with all of these cases. Specifically, it appears that
$$ \det \mathcal{H}_f = (n-1)(-1)^{2n+1} \left[ f(x_1, \dots, x_n) \right]^{n-2} $$
Is this equation true for all $n \in \mathbb{N}_{>1}$?

Assume that $x_j$ are all nonzero (the result is easy to check otherwise).
Using Kronecker delta and $\det\{c_i a_{ij}\}=\det\{c_j a_{ij}\}=(\prod c_j)\det\{a_{ij}\}$, $$\det\mathcal{H}_f=\det_{1\leqslant i,j\leqslant n}\{(1-\delta_{ij})x_i^{-1}x_j^{-1}f(x_1,\ldots,x_n)\}=[f(x_1,\ldots,x_n)]^{n-2}\det(\mathbf{1}_n-\mathbf{I}_n),$$ where $\mathbf{1}_n$ is the all-ones matrix (and $\mathbf{I}_n$ is the identity matrix), which has eigenvalues $0$ (of multiplicity $n-1$) and $n$ (of multiplicity $1$), hence $\det(\lambda\mathbf{I}_n-\mathbf{1}_n)=\lambda^{n-1}(\lambda-n)$. Thus $$\det\mathcal{H}_f=(-1)^{n-1}(n-1)[f(x_1,\ldots,x_n)]^{n-2}.$$
