I am not sure what can be described as unusual. Unlike most other saddle points that I have seen, which have two intersecting lines, these have three. I have also observed no saddle point in system $(b)$ is different from any of the other saddle points. All surrounding trajectories behave in the same way. The saddle points are not identical to the one in $(a)$, but are rotated by $\pi/6$ radians.

In Calculus II only non degenerate stationary points were considered: Hessian (if $n=2$ it is $\begin{pmatrix} H_{xx} &H_{xy} \\ H_{yx} & H_{yy}\end{pmatrix}$) should have it determinant different from $0$ and if it is positive we get extremumâ€”minimum as $H_{xx}>0$ and maximum as $H_{xx}<0$; $H_{yy}$ automatically has the same signâ€”and if determinant is negative we get a saddle).

Here critical point $(0,0)$ is degenerate (and even all second order derivatives vanish here) and it is not the standard saddle. In fact it is called

**Monkey saddle** (three valleys and three mountains joint here while in the usual saddle two valleys and two mountains join). While ordinary saddle is represented by $\renewcommand{\Re}{\operatorname{Re}}$ $\Re (x+iy)^2=x^2-y^2$ the monkey saddle by $\Re (x+iy)^3=x^3-3xy^2$. The multiplicity of the types and complexity of degenerate critical points even as $n=2$ is a reason why you have not looked at them in Calculus II.

In fact system (b) is also integrable for any $\alpha\ne 0$. May be centers are difficult to find directly but there are simple geometric observations.

The beauty of $\alpha=\sqrt{3}=\tan (\pi/3)$ is that triangles are regular (and near critical points it `almost' coincides with (a)). The sides of triangles are $\pi/\sqrt{3}$ so the whole plane is tiled by regular triangles and their vertices are monkey saddles and the whole picture is not only periodic but has a rotational (by $\pi/3$) symmetry and 6 mirror symmetries. Then the rest of stationary points (centers) must be centers of these triangles (and in the regular triangle there is a single center). Here I ignore directions, just purely geometric picture.

Obviously for $\alpha \ne \sqrt{3}$ we need just scale with respect to $x$, so we get equilateral triangles with the base $\pi/\alpha$ and a height $\pi/2$ (the same as before) and the rest of stationary points (centers) must be on the distance $\pi/6$ from the base and they are centroids (points of intersection of medians) of those triangles. Periodic structure remains but only 2 mirror symmetries and rotational by $\pi$ symmetry.

**PS** Up to a factor

\begin{equation}

H(x,y)=\sin(4y)-\sin( 2y+2\alpha x) -\sin(2y-2\alpha x),

\end{equation}

Then as $\alpha=\sqrt{3}$

\begin{equation}

H(x,y)=\sin(4\mathbf{x}\cdot \mathbf{e}_1 )+\sin(4\mathbf{x}\cdot \mathbf{e}_2 ) +\sin(4\mathbf{x}\cdot \mathbf{e}_3)

\end{equation}

with $\mathbf{x}=\begin{pmatrix}x\\y\end{pmatrix}$, $\mathbf{e}_j=\begin{pmatrix}\sin (2\pi (j-1)/3)\\ \cos (2\pi (j-1)/3)\end{pmatrix}$; those are three unit vectors with angles $2\pi/3$ between. Then rotation by $2\pi/3$ preserves $H(x,y)$ and by $\pi/3$ replaces $H(x,y)$ by $-H(x,y)$ which also preserves geometry.