 Solution

The Lagrange multiplier equations are $2x=2\lambda x$, $-2y=2\lambda y$, together with $x^2+y^2=1$. If $x\ne 0$, then $\lambda =1$ and $y=0$; if $y\ne 0$, then $\lambda =-1$ and $x=0$. ThusThe constraint is compact and its gradient never vanishes, so these exhaust the possible global extrema. Equivalently, substituting $x=\cos \theta$, $y=\sin \theta$ gives $z=\cos 2\theta$ along the boundary.