 Solution

An equilibrium has $y=-x/2$ from the first equation. Substitution in the second gives $-x(x^2+3)/2=0$, so the origin is the only real equilibrium. Its Jacobian matrix isPurely imaginary linear eigenvalues alone do not classify the nonlinear equilibrium. Use insteadIts derivatives are $H_x=2x+y+x^3/2$ and $H_y=x+2y$, so the vector field is $(H_y,-H_x)$. The chain rule gives $\dot{H}=H_x{H}_y-H_y{H}_x=0$. Therefore the nonstationary trajectories in the phase portrait lie on the level curvesThe quadratic part is positive definite, with eigenvalues $1/2$ and $3/2$, so $H$ has a strict local minimum at zero. Small positive level sets are smooth closed ovals surrounding it, and the vector field never vanishes on them. They are periodic orbits. Thus the origin is a nonlinear centre, Lyapunov stable but not asymptotically stable. The near-origin ellipses have their long axis along $y=-x$. At $(x,0)$ with $x>0$, the velocity points downward, fixing clockwise orientation.