= Solution
At a <critical point> $p$ of a twice <continuously differentiable> real function, $\nabla f(p)=0$ and the <Taylor expansion> is
$$
f(p+h)=f(p)+\tfrac12h^THh+o(|h|^2),
\qquad H=D^2f(p).
$$
The real symmetric <Hessian matrix> has an <orthonormal eigenbasis>, so writing $h=\sum_i a_i u_i$ gives $h^THh=\sum_i\lambda_i a_i^2$. If every <eigenvalue> is positive, this quadratic form is at least $\lambda_{\min}|h|^2$, and the remainder cannot change its sign for sufficiently small nonzero $h$. Thus $p$ is a strict <local minimum>. If every <eigenvalue> is negative, the analogous argument gives a strict <local maximum>. If positive and negative <eigenvalues> both occur, displacement along the corresponding <eigenvectors> gives values above and below $f(p)$, so $p$ is a <saddle point of a scalar function>.
The <second-derivative test> can be inconclusive when the <Hessian matrix> is singular and semidefinite: higher-order terms along directions in the <kernel of a linear map> matter. For example $x^2+y^4$ and $x^2-y^4$ have the same Hessian $\operatorname{diag}(2,0)$ at zero, but the first has a strict minimum and the second a <saddle point of a scalar function>. A zero <eigenvalue> does not prevent a saddle conclusion if other <eigenvalues> already have both signs.
For the given function write $r^2=x^2+y^2$. Its first <derivatives> are
$$
f_x=y e^{-\alpha r^2}(1-2\alpha x^2),\qquad
f_y=x e^{-\alpha r^2}(1-2\alpha y^2).
$$
The origin is a <critical point> for every real $\alpha$, with
$$
D^2f(0,0)=\begin{pmatrix}0&1\\1&0\end{pmatrix}.
$$
Its <eigenvalues> are $1,-1$, so it is always a <saddle point of a scalar function>. If one coordinate at a <critical point> is zero, the other must also be zero. If both are nonzero, the <derivative> equations force $x^2=y^2=1/(2\alpha)$, possible only for $\alpha>0$. Hence for $\alpha\leq0$ the origin is the only <critical point>.
For $\alpha>0$ there are four additional <critical points>,
$$
(x,y)=\frac1{\sqrt{2\alpha}}(\varepsilon,\eta),
\qquad \varepsilon,\eta\in\{-1,1\}.
$$
At such a point,
$$
f=\frac{\varepsilon\eta}{2\alpha e},\qquad
D^2f=-4\alpha xy\,e^{-1}I_2
=-\frac{2\varepsilon\eta}{e}I_2.
$$
Thus \b[the two equal-sign points are strict maxima, and the two opposite-sign points are strict minima]. They are global: $2|xy|\leq r^2$ implies
$$
|f(x,y)|\leq\frac{r^2}{2}e^{-\alpha r^2}
\leq\frac1{2\alpha e},
$$
where the last maximum occurs at $r^2=1/\alpha$ and equality also requires $x^2=y^2$.
As $\alpha\downarrow0$, the four nonzero <critical points> have distance $\alpha^{-1/2}$ from the origin and \b[escape to infinity], rather than merging with the origin. Their values diverge as $\pm1/(2e\alpha)$. As $\alpha\uparrow0$ from below there are no extra <critical points>, while the origin stays a saddle throughout. This is <critical points escaping to infinity in a Gaussian-weighted bilinear function>.
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