Necessary and sufficient condition for argmax/argmin Let $x_1,\dots,x_n$ be real variables and let $f : \mathbb{R}^n \rightarrow \mathbb{R}$ be a twice-differentiable function with a unique strict global maximum and with a unique maximizer $\mathbf{x}^* = (x_1^*,\dots,x_n^*)$. Assume that $f(x_1,\dots,x_n) = \prod_{i=1}^n f_i(x_i)$, where $f_i$ are twice-differentiable functions. Is there a necessary and sufficient condition for which
$$
(*) \quad \arg \max_{x_1,\dots,x_n} f(x_1,\dots,x_n) = \{ \arg \max_{x_1} f_1(x_1), \dots, \arg \max_{x_n} f_n(x_n) \}
$$
is valid? The same question could be rephrased for $\arg \min$.
 A: A sufficient condition and that is if all the $f_i$ are positive and have a strict unique global maximum. 
We define $x_i^\star$ as the strict unique global maximum of $f_i$ if $f_i(x_i) < f_i(x_i^\star)$ for all $x_i \neq x_i^\star$. Then for a  factorizable  $f(x_1,\ldots,x_n) = \prod_{i=1}^nf_i(x_i)$ and $\forall_i : x_i \neq x_i^\star, $ and we have
$$f(x_1,\ldots,x_n) = f_1(x_1)\cdots f_n(x_n) < f_1(x_1^\star)f_2(x_2)\cdots f_n(x_n) < \cdots < f_1(x_1^\star)\cdots f_n(x_n^\star).$$
While on the other hand we have $f(x_1^\star,\ldots,x_n^\star) = f_1(x_1^\star)\cdots f_n(x_n^\star)$. 
Thus $(x_1^\star,\ldots,x_n^\star)$ is the strict global maximum of $f$.

EDIT: Another sufficient condition is by differentiability. Assume that $f > 0$, then, for any $x_i$, we have that $f_i(x_i) \neq 0$, thus it is or strictly positive or strictly negative. 
Say that $\hat{x}_i$ is the unique maximum of $f_i$, that is $\frac{\partial}{\partial x_i}f_i(\hat{x}_i) = 0$, and $\frac{\partial}{\partial x_i}f_i(x_i) \neq  0$ for $x_i \neq \hat{x}_i$. Also, assume second order differentiability at $\hat{x}_i$, such that $\frac{\partial^2}{\partial x_i^2} f_i(\hat{x}_i) < 0$ by the second derivative test.
Then, by Fermat's theorem, 
$$\begin{matrix}f_2(\hat{x}_2)\cdots f_n(\hat{x}_n)  \cdot \frac{\partial}{\partial x_1} f_1(\hat{x}_1) = 0, \\ \vdots \\ f_1(\hat{x}_1)\cdots f_{n-1}(\hat{x}_{n-1})  \cdot  \frac{\partial}{\partial x_n} f_n(\hat{x}_n) = 0, \\  \end{matrix}$$
Thus, $\hat{x} = (\hat{x}_1,\ldots,\hat{x}_n)$ is a stationary point of $f$. By assumption, it is also the only stationary point of $f$. To see if it is a maximum, the Hessian matrix should have negative eigenvalues. Due to the factorization, the Hessian at $\hat{x}$ is a diagonal matrix, with the $i^{th}$ diagonal entry given by, $$h_{ii} = \left(\frac{\partial^2}{\partial x_i^2} f_i(\hat{x}_i)\right)\cdot \prod_{j \neq i} f_j(\hat{x_j}),$$
and we need $h_{ii} < 0$ for all $i$. We assumed that the second order derrivative of $f_i$ is negative, if $\prod_{j \neq i} f_j(\hat{x_j}) > 0$, we find the result that $x^\star = \hat{x}$ is the global maximum of $f$.

This reflects the comments mady bo OP, that there are different factorizations possible, especially in interchanging the sign. In $h_{ii}$ we see that sign of $f_i$ does not matter, since we only care about the sign of the product at the position of the global optimum. 
