= Solution
At an interior <equilibrium point>, $x^2=1-y^2=\mu-y$, so
$$
y^2-y+\mu-1=0,\qquad \boxed{y_\pm=\frac{1\pm\sqrt{5-4\mu}}2,\quad x_\pm=\sqrt{1-y_\pm^2}.}
$$
For $1<\mu<5/4$ both values satisfy $0<y<1$, giving two physical equilibria. At such an equilibrium,
$$
J=\begin{pmatrix}-2x^2&-2xy\\-2xy&-y\end{pmatrix},\quad \operatorname{tr}J=-2x^2-y<0,\quad \det J=2x^2y(1-2y).
$$
The branch $y_-<1/2$ is <asymptotically stable> by the <trace-determinant stability criterion>; $y_+>1/2$ is a <saddle equilibrium>. In fact $J$ is symmetric, so the stable branch has two negative real eigenvalues.
At $\mu=5/4$ the equilibria merge at $(x,y)=(\sqrt3/2,1/2)$. Writing $y=1/2+s$, the equilibrium condition is $s^2+\mu-5/4=0$, the characteristic branch geometry of a <saddle-node bifurcation>. The stable node and saddle annihilate as $\mu$ increases through $5/4$.
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