= Solution
Write $z=x+y$. When $\mu_1=\mu_2=0$ the system gives $\dot z=y-z^3$ and $\dot y=-z^3$. Choosing
$$
\boxed{V(x,y)=\frac14(x+y)^4+\frac12y^2}
$$
makes $V$ <positive definiteness of a Lyapunov function>[positive definite] and <radially unbounded function>[radially unbounded], while its <orbital derivative> is
$$
\dot V=z^3(y-z^3)-yz^3=-z^6\leq0.
$$
The set $\{\dot V=0\}$ is $z=0$. A trajectory remaining there must also have $\dot z=y=0$, so its largest invariant subset is the origin. The <LaSalle invariance principle> therefore proves that the origin is \b[globally asymptotically stable]. In the <phase plane>, $\dot x=0$ on $y=0$ and $\dot y=0$ on $x+y=0$; every nonstationary trajectory crosses the nested level curves of $V$ inward and tends to the origin.
At an <equilibrium point>, $y=0$ and
$$
x(\mu_1-x^2)=0.
$$
Hence the origin always exists, while $x=\pm\sqrt{\mu_1}$ exist for $\mu_1>0$. At $(x_*,0)$ the <Jacobian matrix> is
$$
J_*=
\begin{pmatrix}
0&1\\
\mu_1-3x_*^2&\mu_2-3x_*^2
\end{pmatrix},
\quad
\det J_*=3x_*^2-\mu_1,
\quad
\operatorname{tr}J_*=\mu_2-3x_*^2.
$$
A <stationary bifurcation> occurs when a real <eigenvalue> passes through zero, hence on
$$
\boxed{\mu_1=0.}
$$
A <Hopf bifurcation> requires zero trace and positive determinant. At the origin this gives
$$
\boxed{\mu_2=0,\quad\mu_1<0,}
$$
and on either nonzero branch it gives
$$
\boxed{\mu_2=3\mu_1,\quad\mu_1>0.}
$$
Thus the Hopf locus consists of the negative $\mu_1$-axis and the ray $\mu_2=3\mu_1$ in the first quadrant; the stationary locus is the $\mu_2$-axis.
Now set $\mu_2=-1$, write $\mu=\mu_1$, and append $\dot\mu=0$. At $(x,y,\mu)=(0,0,0)$ the <extended centre manifold for a parameter> is tangent to the $(x,\mu)$-plane, so put $y=h(x,\mu)$. Its <centre-manifold invariance equation> is
$$
h_xh=\mu x-h-(x+h)^3.
$$
Under the ordering $\mu=O(x^2)$, solving through cubic order gives
$$
\boxed{y=h(x,\mu)=\mu x-x^3+O(x^5),
\qquad
\dot x=\mu x-x^3+O(x^5).}
$$
This is the <pitchfork bifurcation normal form>: $x=0$ is stable for $\mu<0$ and unstable for $\mu>0$, while the two stable branches $x=\pm\sqrt\mu$ emerge for $\mu>0$. The bifurcation is therefore a \b[supercritical pitchfork bifurcation].
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