= Solution
The Jacobian is
$$
J(x,y)=
\begin{pmatrix}
\mu-y^2-6x^2&-2xy\\
-2xy&1-x^2-3y^2
\end{pmatrix}.
$$
At $(0,1)$,
$$
J(0,1)=\operatorname{diag}(\mu-1,-2),
$$
so one eigenvalue crosses zero at
$$
\boxed{\mu=1}
$$
while the other remains negative. Thus a bifurcation occurs there.
On the $x$-axis the nonzero fixed point is
$$
(x,y)=\left(\sqrt{\frac\mu2},0\right).
$$
Its eigenvalues are
$$
-2\mu,\qquad1-\frac\mu2.
$$
The second crosses zero at
$$
\boxed{\mu_c=2},
$$
which is the second bifurcation value.
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