The answer is... close to 1

Using Lagrange multipliers:

Set $f(\mathbf{x}) = x_1x_2x_3x_4x_5x_6, \quad g(\mathbf{x}) = x_1+x_2+x_3+x_4+x_5+x_6, \quad h(\mathbf{x}) = x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2$ There are constants $\lambda, \mu$ such that $\frac{x_1x_2x_3x_4x_5x_6}{x_i} = f_{x_i}(\mathbf{x}) = \lambda g_{x_i}(\mathbf{x}) + \mu h_{x_i}(\mathbf{x}) = \lambda + 2\mu x_i, \text{ for all } 1 \leq i \leq 6.$ Multiplying both sides by $x_i$ and summing over all $i$ , we have $6x_1x_2x_3x_4x_5x_6 = \lambda\left(x_1+x_2+x_3+x_4+x_5+x_6\right) + 2\mu\left(x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2\right) = \lambda\cdot 0 + 2\mu\cdot 6$ so that $x_1x_2x_3x_4x_5x_6 = 2\mu.$ Substituting this back into the Lagrange equations, we have $2\mu = \lambda x_i + 2\mu x_i^2 \\ \Rightarrow x_i^2 + \frac{\lambda}{2\mu}x_i - 1 = 0.$ This means that $x_i$ is one of the two roots of this quadratic, whose values only depend on $\lambda$ and $\mu$ , and since every $x_i$ satisfies the same quadratic, it follows that the list $x_1, x_2, x_3, x_4, x_5, x_6$ contains at most two distinct values. Further, that same quadratic equation also tells us that the two roots are negative reciprocals of one another, so we can summarize all this with $\{x_1,x_2,x_3,x_4,x_5,x_6\} \subseteq \{a, -1/a\}$ for some constant $a$ .

Suppose $k$ of the $x_i$ are equal to $a$ (where by symmetry, we may assume $k \leq 3$ ). Then the first constraint becomes $x_1+x_2+x_3+x_4+x_5+x_6 = 0 \\ \Rightarrow ka + (6-k)\left(-\frac{1}{a}\right) = 0 \\ \Rightarrow a = \pm \sqrt{\frac{6-k}{k}}$ and the value we're trying to maximize becomes $\begin{aligned} f(\mathbf{x}) &= \left(a\right)^k\left(-\frac{1}{a}\right)^{6-k} \\ &= \left(\pm \sqrt{\frac{6-k}{k}}\right)^k\left(\mp \sqrt{\frac{k}{6-k}}\right)^{6-k} \\ &= (-1)^k\left(\frac{k}{6-k}\right)^{3-k}. \end{aligned}$ At this point, we can simply evaluate this expression at all the valid values of $k$ : $\begin{aligned} k=0 &\Rightarrow &\text{Undefined (note that } &a \text{ is undefined for } k=0 \text{)} \\ k=1 &\Rightarrow &-\left(\frac{1}{5}\right)^2 = -\frac{1}{25} \\ k=2 &\Rightarrow &\left(\frac{2}{4}\right)^1 = \frac{1}{2} \\ k=3 &\Rightarrow &-\left(\frac{3}{3}\right)^0 = -1 \end{aligned}$ Of these, the maximum is $\boxed{1/2}$ , so it is the answer.

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A couple technicalities: There are actually two points I skimmed over in the above that I need to cover to actually make this proof formal:

(1) We should keep in mind that this method only finds the absolute maximum under the assumption such an absolute max exists, and I didn't overtly state why we know the max even exists. To guarantee such a maximum exists, we note that the solution set for $x_1+x_2+x_3+x_4+x_5+x_6 = 0$ is closed and the solution set for $x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2 = 6$ is both closed and bounded, so the intersection is both closed and bounded. Since additionally $f(\mathbf{x})$ is continuous, it follows by the Multidimensional Extreme Value Theorem that $f$ attains both a maximum and a minimum.

(2) While the above argument defined the candidate points using the first constraint and therefore we directly see that such points satisfy the first constraint, the second constraint only showed up implicitly through the Lagrange equations, making it much harder to see how we know the second constraint is satisfied. At this point, however, it is easy to check: $k=2 \Rightarrow a=\sqrt{2} \Rightarrow h(a, a, -1/a, -1/a, -1/a, -1/a) = 2 \cdot \left(\sqrt{2}\right)^2 + 4 \cdot \left(-1/\sqrt{2}\right)^2 = 2\cdot 2 + 4\cdot (1/2) = 6$ which explicitly shows the second constraint is also satisfied for such a candidate point.