= Solution
For a positive homogeneous equilibrium,
$$
v=1+u,\qquad \alpha u=\beta v,
$$
so
$$
u_* =\frac\beta{\alpha-\beta},\qquad
v_* =\frac\alpha{\alpha-\beta},
$$
requiring $\alpha>\beta$. The reaction Jacobian there is
$$
J=\frac1{\alpha-\beta}
\begin{pmatrix}
\beta&-\beta\\
\alpha\beta&-\beta^2
\end{pmatrix}.
$$
Its <determinant> is $\beta^2/(\alpha-\beta)>0$, while its trace is $\beta(1-\beta)/(\alpha-\beta)$. Stability therefore requires
$$
\beta>1,\qquad \alpha>\beta.
$$
This is the region above the line $\alpha=\beta$ and to the right of $\beta=1$ in the positive quadrant.
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