= Solution
The constraint set $S^1\times S^2$ is <compact>, so the continuous objective attains a maximum and minimum. The two constraint <gradients> are nonzero and supported on separate coordinates, so they are <linearly independent> everywhere on it. The <Lagrange multiplier> equations are
$$
y_2-y_3=2\lambda x_1,\quad0=2\lambda x_2,\quad0=2\mu y_1,\quad x_1=2\mu y_2,\quad-x_1=2\mu y_3.
$$
Both positive and negative values occur. At a maximizing or minimizing point the objective is therefore nonzero, so $x_1\ne0$, $y_2-y_3\ne0$, and both $\lambda,\mu$ are nonzero. The equations imply $x_2=0$, $y_1=0$, and $y_3=-y_2$. The constraints then give $x_1=\pm1$, $y_2=\pm1/\sqrt2$.
Consequently, writing $s=\pm1$, the complete list of extrema is
$$
\boxed{\max=\sqrt2:\ (x_1,x_2;y_1,y_2,y_3)=(s,0;0,s/\sqrt2,-s/\sqrt2)},
$$
$$
\boxed{\min=-\sqrt2:\ (x_1,x_2;y_1,y_2,y_3)=(s,0;0,-s/\sqrt2,s/\sqrt2)}.
$$
For an independent global check, <Cauchy-Schwarz inequality> gives $|x_1(y_2-y_3)|\leq|x_1|\sqrt{2(y_2^2+y_3^2)}\leq\sqrt2$, with equality exactly at the displayed points.
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