You were doing well until the very end. Note that the determinant of the Hessian, which is also a multiple of the discriminant of the original quadric, vanishes precisely when $a=\pm2$, so this case is really the same as the next one that you describe, and the critical point at the origin is in fact also indeterminate.
To determine the nature of the critical points in this case, it’s helpful to step back from the calculus and algebra and take a geometric look at the situation. When $|a|\lt 2$, the graph of this function is an elliptic paraboloid opening upward with vertex at the origin. For $|a|\gt2$, we have a classic saddle surface. Finally, when $|a|=2$, we have a parabolic cylinder, with the bottom of the “valley” along the line $x+y=0$ or $x-y=0$. So, in the indeterminate case, the function is constant along this line and increasing in every other direction away from the line.
You can examine the spectrum of the Hessian matrix to determine all of this: at a critical point, the first-order derivatives vanish, so per its Taylor expansion, the behavior of the function near this point is dominated by the quadratic form determined by the second derivatives. In other words, near a critical point, the function “looks like” the quadric defined by its Hessian. The signs of the Hessian’s eigenvalues thus tell you whether the function is generally increasing or decreasing in the principal axis directions near a critical point.
For a $2\times2$ Hessian, a positive determinant means that the eigenvalues are either both positive—a local minimum—or both negative—a local maximum. You can examine the sign of its trace, or more simply, its upper-left entry, to determine which of these two cases you have. A negative determinant means that the eigenvalues have opposite signs—increasing in one principal direction but decreasing in the other, i.e., a saddle point. A zero determinant means that there’s at least one zero eigenvalue, so we have to probe more deeply.
For this particular family of functions, the trace of the Hessian is always positive, so when the determinant vanishes we have one positive and one zero eigenvalue: the function is increasing along one principal axis. For its behavior in the other principal direction, we have to look at higher-order derivatives or examine the behavior of the function near the critical point directly. In this case, we’ve run out of derivatives—all third- and higher-order derivatives vanish—but it’s easy to see that the function is zero all along the “trough” of the cylinder, as you’ve determined.
Best Answer
First, what you found to be $\sqrt2/\mathrm e^2$ is the norm of the gradient vector, not the gradient vector. Second, the exercise is very badly phrased; a quadrant doesn't "represent" a local maximum or minimum, it contains one. Third, you can't directly find minima and maxima from individual gradient vectors, but in the present case the idea is to use the level curves to see that there are extrema near $(\pm0.7,\pm0.7)$ and then use the gradients at $(\pm1,\pm1)$ to decide which of those extrema are minima or maxima by finding whether the gradient points towards or away from the extremum.