= Solution
Use the homogeneous <equilibrium point> condition $k(\rho_0)\rho_0=a_0f(\rho_0)$. Evaluate $D_1,D_2$ at $(a_0,\rho_0)$ and abbreviate
$$
f_0=f(\rho_0),\quad f'_0=f'(\rho_0),\quad
\kappa=k(\rho_0)+\rho_0k'(\rho_0),\quad
c=\kappa-a_0f'_0.
$$
The quantity $c$ is the chemical relaxation rate with the cell density held fixed. Derivatives of $D_1,D_2$ do not enter this <linear stability analysis>: they multiply spatial derivatives of the homogeneous background or products of perturbations. For a <Fourier mode> with time dependence $e^{st}$, put $z=|\mathbf q|^2$ and write
$$
s\binom{\widehat a}{\widehat\rho}=M(z)\binom{\widehat a}{\widehat\rho},\qquad
M(z)=\begin{pmatrix}-D_2z&D_1z\\ f_0&-c-D_\rho z\end{pmatrix}.
$$
Consequently
$$
\operatorname{tr}M=-c-(D_2+D_\rho)z,\qquad
\det M=z[D_2(c+D_\rho z)-D_1f_0].
$$
At $z=0$, the <eigenvalues> are $0$ and $-c$. The neutral cell-density mode reflects <mass conservation>, not decay of every homogeneous perturbation.
Write $a=D_2>0$, $b=D_\rho>0$, $C=D_1f_0$, and interpret production physically as $f_0\geq0$. If $c\geq0$, the <trace-determinant stability criterion> shows that an unstable nonzero <wavenumber> exists precisely when $C>ac$. The unstable band is
$$
0<z<z_c,\qquad z_c=\frac{C-ac}{ab}.
$$
If $c<0$, the chemical field is already unstable at zero <wavenumber>; arbitrarily small positive <wavenumbers> are unstable too. Since $C\geq0$, this also implies $C>ac$. Thus, on the infinite plane, \b[the undivided <Keller--Segel aggregation threshold> is]
$$
\boxed{D_1f_0>D_2(\kappa-a_0f'_0).}
$$
For $\kappa>0$, dividing by $D_2\kappa$ gives the requested form
$$
\boxed{\frac{D_1f_0}{D_2\kappa}+\frac{a_0f'_0}{\kappa}>1.}
$$
At equality there is no strictly growing mode when $c\geq0$; nonzero spatial modes decay. On a finite domain, a permitted nonzero <wavenumber> must actually lie in the unstable band. If one allows a signed production function, the displayed matrix and <trace-determinant stability criterion> remain valid, but the simplification using $C\geq0$ must be revisited.
\b[The printed hypotheses do not ensure $\kappa>0$.] Positivity of the degradation rate $k$ alone is insufficient. For a concrete counterexample, take $k(\rho)=e^{-2\rho}$, $f(\rho)=e^{-2}$, $a_0=\rho_0=1$, and $D_1=D_2=D_\rho=1$. The homogeneous <equilibrium point> condition holds, all rates and transport coefficients are positive, but $\kappa=c=-e^{-2}$. The printed left-hand side equals $-1$, although $\det M=z(z-2e^{-2})<0$ for $0<z<2e^{-2}$. These spatial modes grow. If $\kappa=0$, the printed expression is undefined. The undivided criterion and the matrix above resolve both cases.
To find the <fastest-growing Keller--Segel mode>, use the larger <eigenvalue>
$$
s_+(z)=-\frac{c+(a+b)z}{2}+\frac12\sqrt{[c+(b-a)z]^2+4Cz}.
$$
For $C>0$ the square root is real. Put $p=a+b$ and $d=b-a$. Differentiation gives
$$
s_+'(z)=-\frac p2+\frac{d(c+dz)+2C}{2\sqrt{(c+dz)^2+4Cz}}.
$$
A positive maximizing <wavenumber> exists in either of two growing cases: $c\geq0$, $C>ac$, or $c<0$, $C>-bc$. Equivalently, $C>\max(ac,-bc)$. In this range $C+dc>0$, and
$$
s_+''(z)=-\frac{2C(C+dc)}{[(c+dz)^2+4Cz]^{3/2}}<0.
$$
Thus the stationary point is the unique maximum. The derivative condition yields
$$
ab[(c+dz_*)^2+4Cz_*]=C(C+dc).
$$
Solving it, with the root that satisfies the unsquared derivative equation, gives
$$
\boxed{z_*=
\begin{cases}
\displaystyle\frac{p\sqrt{C(C+dc)/(ab)}-dc-2C}{d^2},&a\ne b,\\[6pt]
\displaystyle\frac{C^2/a^2-c^2}{4C},&a=b.
\end{cases}\qquad
\ell_*=\frac{2\pi}{\sqrt{z_*}}.}
$$
For $c=0$, the unequal-diffusivity expression simplifies to $z_*=C/[\sqrt{ab}(\sqrt a+\sqrt b)^2]$, also agreeing with the equal-diffusivity limit. The formula maximizes the full two-field growth rate; it makes no instantaneous-chemical approximation. Every direction of $\mathbf q$ with this length is equivalent by <rotational symmetry>.
If $c<0$ but $0\leq C\leq-bc$, the maximum instead occurs at $z=0$: \b[the fastest mode is homogeneous, with $s_*=-c$ and infinite <wavelength>]. For $C>0$, its initial derivative is $-b+C/(-c)\leq0$. If $C+dc\geq0$ the derivative thereafter decreases, while if $C+dc<0$ it increases towards the still-negative large-$z$ limit; either way no positive-$z$ maximum is missed. At $C=0$, the two <eigenvalues> are simply $-az$ and $-c-bz$, giving the same conclusion. In a finite box, maximize $s_+$ over the allowed <Fourier modes>; excluding the homogeneous mode can change the selected length. In a stable parameter range there is no fastest-growing mode.
The first ratio compares the positive feedback loop “more cells produce more attractant, which draws in more cells” with spreading by cell <diffusion> and removal of chemical perturbations. Its numerator $D_1f_0$ measures <chemotaxis> together with attractant production; its denominator $D_2\kappa$ measures dispersal together with incremental degradation. \b[This competition between directed <chemotaxis> and cell <diffusion> produces spatial aggregation.] The second ratio compares the concentration dependence of chemical production, $a_0f'_0$, with incremental degradation $\kappa$. Positive $f'_0$ amplifies chemical fluctuations, whereas negative $f'_0$ suppresses them. It changes the chemical relaxation available to the <chemotaxis> feedback loop and can itself destabilize the homogeneous chemical field. Chemical <diffusion> $D_\rho$ suppresses short scales and sets the selected <wavelength>, but does not change the infinite-plane long-wave threshold. These interpretations as ratios of stabilizing and destabilizing processes assume $\kappa>0$.
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