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The fixed points are $(0,0)$ and $(\pm |\mu |,{\mu }^2)$. The Jacobian matrix isAt the origin its eigenvalues are ${\mu }^2$ and $-1$, so it is a saddle equilibrium for every allowed $\mu$. At either nonzero fixed point the characteristic polynomial is ${\lambda }^2+\lambda +2{\mu }^2$, givingThus both are stable nodes for $0<|\mu |<1/(2\sqrt{2})$ and stable focuses for $|\mu |>1/(2\sqrt{2})$. At equality the repeated eigenvalue is $-1/2$ and the matrix is not scalar, so there is only one independent eigenvector: the fixed point is a stable improper node.

The stable manifold of the origin is $x=0$, exactly: there $\dot{y}=-y$. Conversely, a trajectory converging to the origin with $x\ne 0$ would eventually have $\dot{x}/x={\mu }^2-y>{\mu }^2/2$, which precludes convergence of $x$ to zero.

For the local unstable manifold, write $y=h(x)=a{x}^2+b{x}^4+O(x^6)$. Reflection symmetry makes $h$ even, and the invariance equation for a graph requiresComparing quadratic and quartic terms gives $2a{\mu }^2=1-a$ and $4b{\mu }^2-2a^2=-b$. Therefore