= Solution
The $u^2/v$ term makes $u$ locally autocatalytic, while increasing $v$ suppresses this production. Thus \b[$u$ is the activator and $v$ is the inhibitor].
At a positive <spatially homogeneous equilibrium>,
$$
\frac{u^2}{v}-2bu=0,
\qquad
u^2-v=0.
$$
Hence $u=2bv$ and $v=u^2$, giving
$$
\boxed{(u_*,v_*)=\left(\frac1{2b},\frac1{4b^2}\right).}
$$
The reaction <Jacobian matrix> there is
$$
J=
\begin{pmatrix}
2b&-4b^2\\
1/b&-1
\end{pmatrix},
\qquad
\operatorname{tr}J=2b-1,
\qquad
\det J=2b.
$$
Because $b>0$, the <trace-determinant stability criterion> shows that the spatially uniform reaction kinetics are stable exactly when
$$
\boxed{0<b<\frac12.}
$$
For a perturbation proportional to $e^{\lambda t+ikx}$, put $q=k^2$. Its growth rates are the eigenvalues of
$$
J-q
\begin{pmatrix}d&0\\0&1\end{pmatrix}.
$$
The trace is $2b-1-(d+1)q<0$ in the reaction-stable regime, while the determinant is
$$
\Delta(q)=dq^2+(d-2b)q+2b.
$$
A <Turing instability> occurs exactly when this upward-opening quadratic is negative for some $q>0$. By the <two-species diffusion-driven instability criterion>, this requires
$$
d<2b,
\qquad
(2b-d)^2>8bd.
$$
Together these reduce to
$$
\boxed{0<b<\frac12,\qquad 0<d<(6-4\sqrt2)b.}
$$
Thus the unstable region in the $(b,d)$-plane lies below the straight line $d=(6-4\sqrt2)b$ and to the left of $b=1/2$.
Inside this region, the unstable squared-wavenumber band is
$$
\boxed{\frac{2b-d-\sqrt{(2b-d)^2-8bd}}{2d}
<k^2<
\frac{2b-d+\sqrt{(2b-d)^2-8bd}}{2d}.}
$$
At the bifurcation boundary the two roots coincide, so
$$
k_c^2=\frac{2b-d}{2d}.
$$
Substituting $d=(6-4\sqrt2)b$ yields
$$
\boxed{k_c=\sqrt{1+\sqrt2}.}
$$
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