= Solution
Write the reaction terms as
$$
f(u,v)=u(1+u-\gamma v),\qquad
g(u,v)=v(\beta u-v).
$$
A nonzero homogeneous equilibrium satisfies $v_*=\beta u_*$ and $1+u_*-\gamma v_*=0$, hence
$$
\boxed{u_*=\frac1{\beta\gamma-1},\qquad
v_*=\frac{\beta}{\beta\gamma-1}}.
$$
For physically positive populations it exists exactly when
$$
\boxed{\beta\gamma>1}.
$$
The reaction <Jacobian matrix> at this equilibrium is
$$
J=u_*
\begin{pmatrix}
1&-\gamma\\
\beta^2&-\beta
\end{pmatrix}.
$$
Its trace and determinant are
$$
\operatorname{tr}J=u_*(1-\beta),\qquad
\det J=u_*^2\beta(\beta\gamma-1)>0.
$$
The equilibrium is therefore stable to spatially uniform perturbations when
$$
\boxed{\beta>1,\qquad\beta\gamma>1}.
$$
For a spatial <Fourier mode> of wavenumber $k$, put $q=k^2$. The linearized <reaction-diffusion system> has matrix
$$
J_q=J-q\begin{pmatrix}1&0\\0&d\end{pmatrix}.
$$
Its trace is smaller than $\operatorname{tr}J$, while
$$
\det J_q
=dq^2-u_*(d-\beta)q
+u_*^2\beta(\beta\gamma-1).
$$
The <two-species diffusion-driven instability criterion> says that this upward-opening quadratic becomes negative for some $q>0$ precisely when
$$
d>\beta,\qquad
(d-\beta)^2>4d\beta(\beta\gamma-1).
$$
Combining all conditions, a <Turing instability> may occur in the region
$$
\boxed{
\beta>1,\qquad d>\beta,\qquad
\frac1\beta<\gamma<
\frac1\beta+\frac{(d-\beta)^2}{4d\beta^2}}.
$$
At onset the discriminant vanishes, and the double root is
$$
q_c=\frac{u_*(d-\beta)}{2d}.
$$
The threshold relation gives
$$
\beta\gamma-1=\frac{(d-\beta)^2}{4d\beta},
\qquad
u_*=\frac{4d\beta}{(d-\beta)^2},
$$
so the critical wavenumber is
$$
\boxed{k_c=\left(\frac{2\beta}{d-\beta}\right)^{1/2}}.
$$
If $d=1$, uniform stability requires $\beta>1$ whereas diffusion-driven instability requires $d>\beta$. These inequalities are incompatible, so the Turing region vanishes.
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