 Solution

On the unit circle the constraint gradient $(2x,2y)$ never vanishes. The Lagrange multiplier equations areA nonzero solution of the first two equations requiresso $\lambda =9/2$ or $\lambda =-1/2$. Multiplying the first equation by $x$, the second by $y$, and adding shows $2f=2\lambda (x^2+y^2)=2\lambda$. Thus each candidate value of $f$ equals its multiplier.

For $\lambda =9/2$, $y=-x/3$, giving $(x,y)=\pm (3,-1)/\sqrt{10}$. For $\lambda =-1/2$, $y=3x$, giving $(x,y)=\pm (1,3)/\sqrt{10}$. Since a continuous function attains its extrema on the compact set given by the unit circle and these exhaust the Lagrange multiplier candidates,Each $\pm$ changes both coordinates together.