**(a)** Since $x^2+y^2+4>0$ all stationary points are at $y=0$, and either $x=0$ or $x^2+y^2-16=0$; thus we have

$A_1=(0,0)$ and $A_{2,3}=(\pm 4,0)$.

**(b)** Let $f=2y(x^2+y^2+4)$, $g=-2x(x^2+y^2-16)$.

At $A_1$ we have $f_x=0$, $f_y=8$, $g_x=32$, $g_y=0$ and linearization is

$$\begin{pmatrix}X\\ Y\end{pmatrix}'=\begin{pmatrix} 0 &8\\32 &0\end{pmatrix}\begin{pmatrix}X\\Y\end{pmatrix}$$

with eigenvalues $\pm 16$ and eigenvectors $\begin{pmatrix}1 \\\pm2\end{pmatrix}$ so it is a saddle:

At points $A_{2,3}$ we have $f_x=0$, $f_y=16$, $g_x=-16$, $g_y=0$ with eigenvalues $16i$ and $-16i$; in virtue of (c) it is a center, and since the bottom left element is negative it is clockwise.

Rewriting the system as

$$

2x(x^2+y^2-16)\,dx + 2y(x^2+y^2+4)\,dy=0

$$

one can check that the form on the left is exact; $H_x= 2(x^3+xy^2-16x)\implies H(x,y)=\frac{1}{2}(x^4 +2x^2y^2 -32 x^2)+h(y)$ and plugging to $H_y=2x^2y +h'(y)= 2yx^2+2y^3 +8y\implies h'(y)=2y^3 +8y\implies h(y)=\frac{1}{2}y^4+4y^2$ (we take a constant equal $0$). Then

$$H(x,y)=\frac{1}{2}\bigl(x^4 +2x^2y^2 +y^4-32 x^2 +8y^2\bigr).$$

Points $(\pm 4,0)$ are minima of $H(x,y)$ and $(0,0)$ is the saddle point. So called symmetric two-well 2D potential.

See 3D plot generated by WolframAlpha