 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 areBoth positive and negative values occur. At a maximizing or minimizing point the objective is therefore nonzero, so $x_1\ne 0$, $y_2-y_3\ne 0$, and both $\lambda ,\mu$ are nonzero. The equations imply $x_2=0$, $y_1=0$, and $y_3=-y_2$. The constraints then give $x_1=\pm 1$, $y_2=\pm 1/\sqrt{2}$.

Consequently, writing $s=\pm 1$, the complete list of extrema isFor an independent global check, Cauchy-Schwarz inequality gives $|x_1(y_2-y_3)|\le |x_1|\sqrt{2(y_2^2+y_3^2)}\le \sqrt{2}$, with equality exactly at the displayed points.