= 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 \b[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 $|f|\leq\sqrt2r/(1+r^2)\leq1/\sqrt2$, with equality at exactly those two points. This is the extremum bound underlying the <level sets of a linear numerator over a radial quadratic>.
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