Let $P=x_1\cdots x_6$. We assume $x_1\ge\cdots\ge x_6$ and $x_4\le0$(if $x_4>0$, we can let $y_1=-x_6,\cdots,y_6=-x_1$, then $y_4=-x_3<0$). So $x_1\ge0$ and $x_6\le x_5\le x_4\le0$. Case 1: $x_2\le0$ We have $P\le0$. Case 2: $x_2>0$ Case 2.1: $x_3\ge 0$ We have $P\le0$. Case 2.2: $x_3<0$ Let $a=x_3+x_4,b=x_5+x_6$. So $$6=x_1^2+\cdots+x_6^2\ge\frac{(x_1+x_2)^2+(x_3^2+x_4)^2+(x_5+x_6)^2}{2}=\frac{(a+b)^2+a^2+b^2}{2}\ge\frac{3(a+b)^2}{4}.$$Hence $(a+b)^2\le8$. We have $$P=(x_1x_2)(x_3x_4)(x_5x_6)\le\frac{1}{4^3}(x_1+x_2)^2(x_3+x_4)^2(x_5+x_6)^2$$$$=\frac{1}{64}(a+b)^2a^2b^2\le\frac{1}{64}(a+b)^2\cdot\frac{(a+b)^4}{16}$$$$\le\frac{1}{64}\cdot8\cdot\frac{8^2}{16}=\frac{1}{2}.$$Equality holds iff $x_1=x_2=\sqrt{2},x_3=\cdots=x_6=-\frac{1}{\sqrt{2}}$, or $x_1=x_2=-\sqrt{2},x_3=\cdots=x_6=\frac{1}{\sqrt{2}}$, or their permutations.