Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 12 / b / Solution

At $(t,t)$ with $t^2=1/2$, differentiating the gradient yields the Hessian matrix $H=-tI$. At the positive point it is negative definite, and at the negative point it is positive-definite; in both cases $f_{xy}=0$ and $\det H=1/2$. Therefore the positive point is a maximum, the negative point is a minimum, and there are no saddle points.

They are global extrema: with $r=\sqrt{x^2+y^2}$, the Cauchy-Schwarz inequality gives $|f|\le \sqrt{2}r/(1+r^2)\le 1/\sqrt{2}$, with equality at exactly those two points. This is the extremum bound underlying the level sets of a linear numerator over a radial quadratic.