= Solution
Along a family of equilibria at fixed temperature, $K$ and the dimensionless function $\Psi$ are fixed. The boundary <mass density> changes through both $\rho_c$ and the boundary coordinate $\xi$:
$$
\boxed{\delta\rho=e^{-\Psi}\delta\rho_c-\rho_ce^{-\Psi}\Psi'\delta\xi
=\rho\left(\frac{\delta\rho_c}{\rho_c}-\Psi'\delta\xi\right).}
$$
From part (c), $N=C\rho_c^{-1/2}f(\xi)$, where $C$ is constant at fixed temperature and $f=\xi^2\Psi'$. Set its logarithmic variation to zero:
$$
0=\frac{\delta N}{N}=-\frac12\frac{\delta\rho_c}{\rho_c}
+\frac{f'}f\delta\xi.
$$
The <isothermal Lane-Emden equation> gives $f'=\xi^2e^{-\Psi}$, so the fixed-$N$ relation is
$$
\boxed{\frac{\delta\rho_c}{\rho_c}=\frac{2e^{-\Psi}}{\Psi'}\delta\xi.}
$$
Substitute back into the boundary-density variation:
$$
\delta\rho\big|_N
=\rho\left(\frac{2e^{-\Psi}}{\Psi'}-\Psi'\right)\delta\xi.
$$
For a nontrivial variation at a finite positive radius, its vanishing gives the <fixed-mass turning point of an isothermal sphere>:
$$
\boxed{[\Psi'(\xi)]^2=2e^{-\Psi(\xi)}.}
$$
This derivative follows a fixed-mass sequence of equilibria; it is not merely the spatial derivative of density inside one sphere. The corresponding radial variation is
$$
\frac{\delta r}{r}=\left(\frac1\xi-\frac{e^{-\Psi}}{\Psi'}\right)\delta\xi.
$$
At small $\xi$, $\Psi'\simeq\xi/3$: the boundary density increases while the radius decreases along increasing $\xi$, giving the stated negative $(\partial p/\partial V)_{N,T}$ of the stable branch. The first density turning point marks its first loss of radial <stability>; it need not be confused with a turning point of radius itself.
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