= Solution
Set
$$
\varepsilon=\mu-1,
\qquad
v=y-1,
\qquad
\dot\varepsilon=0.
$$
Then
$$
\begin{aligned}
\dot x&=x(\varepsilon-2v-v^2-2x^2),\\
\dot v&=-2v-x^2-vx^2-3v^2-v^3.
\end{aligned}
$$
The extended centre variables are $(x,\varepsilon)$ and the stable variable is $v$. Write the centre manifold as $v=h(x,\varepsilon)$. At quadratic order, the invariance equation
$$
h_x\dot x+h_\varepsilon\dot\varepsilon
=-2h-x^2-vx^2-3v^2-v^3
$$
gives
$$
h(x,\varepsilon)=-\frac{x^2}{2}+\text{higher-order terms}.
$$
Substitution into the $x$ equation gives the leading normal form
$$
\boxed{\dot x=\varepsilon x-x^3+\text{higher-order terms}.}
$$
This is a supercritical pitchfork: the $x=0$ branch loses stability as $\varepsilon$ becomes positive and stable branches $x=\pm\sqrt\varepsilon$ emerge. The original system is invariant under $x\mapsto-x$, so a pitchfork is exactly the symmetry-forced bifurcation expected. The physical quadrant $x\geq0$ displays its positive half.
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