= Solution
At a <critical point> $x_0$ of a twice continuously differentiable real function, the <Taylor expansion> is
$$
f(x_0+h)=f(x_0)+\frac12h^THh+o(|h|^2),\qquad H=\nabla^2f(x_0).
$$
The <Hessian matrix> is real symmetric, so an orthogonal eigenbasis diagonalizes its <quadratic form>. If every <eigenvalue> is positive, that form is bounded below by a positive multiple of $|h|^2$, which dominates the remainder and gives a strict <local minimum>. If every <eigenvalue> is negative, the same argument gives a strict <local maximum>. If both signs occur, displacement along the corresponding eigenvectors produces values above and below $f(x_0)$, so the point is a <saddle point of a scalar function>. These exhaust a nonsingular <Hessian matrix>.
A singular <Hessian matrix> does not always make the classification impossible: nonzero <eigenvalues> of both signs still give a saddle. If the <Hessian matrix> is positive or negative semidefinite, higher-order terms in its null directions must be examined. For example, $x^2+y^4$ and $x^2-y^4$ have the same <Hessian matrix> $\operatorname{diag}(2,0)$ at zero, but the first has a strict minimum and the second a saddle. The function $x^2$ has a non-strict minimum along a whole line.
For the given polynomial,
$$
f_x=2x(\alpha-3y+4x^2),\qquad f_y=2y-3x^2.
$$
A stationary point has $y=3x^2/2$ and then $x(2\alpha-x^2)=0$. Hence the origin is always critical, and two additional points exist exactly for $\alpha>0$:
$$
\boxed{(x,y)=(\pm\sqrt{2\alpha},3\alpha).}
$$
The <Hessian matrix> is
$$
H=\begin{pmatrix}2\alpha-6y+24x^2&-6x\\-6x&2\end{pmatrix}.
$$
At the origin it is $\operatorname{diag}(2\alpha,2)$, giving a strict <local minimum> for $\alpha>0$ and a saddle for $\alpha<0$. At either extra <critical point> it has <determinant> $64\alpha-36(2\alpha)=-8\alpha<0$, so both extra points are saddles.
At $\alpha=0$, complete the square:
$$
f(x,y)=\left(y-\frac32x^2\right)^2-\frac14x^4.
$$
Along $x=0$ it is positive away from zero, whereas along $y=3x^2/2$ it is negative away from zero. Thus the origin remains a degenerate saddle despite its nonnegative <Hessian matrix>. This is the <parabolic quartic critical-point classification>. The complete result is
$$
\boxed{\begin{array}{c|c}
\alpha<0&(0,0)\text{ is the only critical point, a saddle}\\
\alpha=0&(0,0)\text{ is the only critical point, a degenerate saddle}\\
\alpha>0&(0,0)\text{ is a strict local minimum; }(\pm\sqrt{2\alpha},3\alpha)\text{ are saddles}.
\end{array}}
$$
For $\alpha>0$ the <local minimum> is not global, since along the same parabola $f=\alpha x^2-x^4/4\to-\infty$.
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