= Solution
For a flow, a <hyperbolic fixed point> has no <eigenvalue> of its <linearization> with zero real part; a <nonhyperbolic fixed point> has at least one such <eigenvalue>. At $(\mu,0)$ the <eigenvalues> are $-\mu,1-\mu$, while at $(0,\pm1)$ they are $\mu+1,-2$. Thus nonhyperbolicity occurs at the two stated parameters. The following <centre manifold> reductions show the associated actual bifurcations.
Near $(1,0)$ put $X=x-\mu$, $Y=y$, $\nu=\mu-1$. The transformed equations are
$$
\dot X=(1+\nu+X)(-X+Y^2),\qquad
\dot Y=Y(-\nu-X-Y^2),\qquad\dot\nu=0.
$$
At $\nu=0$, the non-extended stable subspace is the $X$ axis and the centre subspace is the $Y$ axis. In the <extended centre manifold>, write $X=h(Y,\nu)$, with $h(0,\nu)=0$. The symmetry $Y\mapsto-Y$ allows an even graph. At quadratic order its invariance equation has left side of degree at least three and right side $-h+Y^2$, so $h=Y^2+O(\nu Y^2,Y^4)$. Substitution gives
$$
\boxed{\dot Y=-\nu Y-2Y^3+O(\nu Y^3,Y^5).}
$$
This is a <supercritical pitchfork bifurcation> in parameter $-\nu=1-\mu$. The central branch is stable for $\mu>1$ and unstable for $\mu<1$ near one; the two nonzero branches for $\mu<1$ are stable. At the bifurcation itself the cubic term attracts on the <centre manifold>.
Near $(0,1)$ put $X=x$, $Y=y-1$, $\nu=\mu+1$. Then
$$
\dot X=X(\nu-X+2Y+Y^2),\qquad
\dot Y=(1+Y)(-X-2Y-Y^2),\qquad\dot\nu=0.
$$
At $\nu=0$, the stable subspace is $X=0$ and the centre line is $Y=-X/2$. Thus the leading extended centre graph is $Y=-X/2+O(X^2,\nu X)$, and the reduced flow is
$$
\boxed{\dot X=\nu X-2X^2+O(X^3,\nu X^2).}
$$
This is a <transcritical bifurcation>: $X=0$ is stable for $\nu<0$, while the branch $X=\nu/2+\cdots$ is stable for $\nu>0$; the opposite branches are unstable. At the critical point the quadratic term is only one-sided attracting.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ii/paper-4-bifurcations.png]
{title=Local pitchfork and transcritical bifurcations, with stable branches solid and unstable branches dashed}
The exact additional branch in $y>0$ is
$$
\boxed{x=(\mu+1)/2,\qquad y=\sqrt{(1-\mu)/2},\qquad\mu<1.}
$$
It meets $(\mu,0)$ at $\mu=1$ and $(0,1)$ at $\mu=-1$, displaying the <connecting pitchfork and transcritical branches in a quadratic-product flow>. Its Jacobian has determinant $1-\mu^2$ and trace $(\mu-3)/2$. Hence it is stable for $-1<\mu<1$ and a saddle for $\mu<-1$. The negative-$y$ companion is needed for the pitchfork sketch near $(1,0)$; no separate analysis of $(0,-1)$ is required.
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