= Solution
\b[As printed, the <function> is undefined at the origin.] For $r^2=x^2+y^2>0$, its <gradient> is
$$
\nabla f=\frac{2e^{-1/r^2}}{r^4}(x,y).
$$
The <scalar> prefactor is positive, and $(x,y)\ne(0,0)$, so this <gradient> never vanishes. Thus there are \b[no <stationary points> on the stated domain]. Its <infimum> is three, approached as $r\to0$, but that value is not attained there.
If the intended definition includes the natural extension $f(0,0)=3$, that extension has a \b[unique strict <global minimum> at the origin], with value three. Indeed, $e^{-1/r^2}>0$ off zero, and $e^{-1/r^2}/r\to0$ proves differentiability at zero with zero <gradient>. More strongly, subtracting three gives the <radial Gaussian flat function>: every <derivative> of that difference vanishes at the origin. The vanishing <Hessian matrix> does not turn this minimum into an <inflection point>. These are two different domain interpretations; the extension must be stated rather than silently inserted.
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