The photon sphere and the innermost stable circular orbit

Step 1 (conserved quantities and the radial equation). Along any geodesic (timelike $\kappa=-c^2$, or null $\kappa=0$) confined to the equatorial plane $\theta=\pi/2$, the metric's independence of $t$ and $\phi$ gives two conserved quantities $\tilde E = (1-r_s/r)c^2\dot t$ and $\tilde L = r^2\dot\phi$ (dot $=d/d\tau$). Substituting these into the normalization $g_{\mu\nu}\dot x^\mu\dot x^\nu=\kappa$ and simplifying yields, for timelike motion, $$\left(\frac{dr}{d\tau}\right)^2 = \tilde E^2 - \left(1-\frac{r_s}{r}\right)\left(c^2+\frac{\tilde L^2}{r^2}\right) \equiv \tilde E^2 - V^2_{\text{eff}}(r)$$

Step 2 (circular-orbit condition). Expanding gives $$V^2_{\text{eff}}(r) = c^2 - \frac{c^2 r_s}{r} + \frac{\tilde L^2}{r^2} - \frac{\tilde L^2 r_s}{r^3}$$ A circular orbit has constant $r$, i.e. $dr/d\tau=0$ at all times, which requires both $\tilde E^2=V^2_{\text{eff}}(r)$ and (so that $r$ does not drift away) $dV^2_{\text{eff}}/dr=0$: $$\frac{d}{dr}V^2_{\text{eff}} = \frac{c^2 r_s}{r^2} - \frac{2\tilde L^2}{r^3} + \frac{3\tilde L^2 r_s}{r^4} = 0$$

Step 3 (solve for the angular momentum of a circular orbit). Multiplying through by $r^4$ and solving for $\tilde L^2$ gives the angular momentum needed to sustain a circular orbit at radius $r$: $$\tilde L^2(r) = \frac{c^2 r_s r^2}{2r-3r_s}$$ This already carries information about the photon sphere: as $r\to (3/2)r_s$ from above, the denominator $2r-3r_s\to0^+$ and $\tilde L^2\to+\infty$ — no finite angular momentum sustains a circular orbit that close, which is exactly the massless (photon) limit reached below.

Step 4 (marginal stability — the ISCO). As $r$ decreases from infinity, $\tilde L^2(r)$ first decreases, reaches a minimum, then blows up at $r=3r_s/2$; a circular orbit is stable exactly where increasing $\tilde L^2$ is needed to shrink $r$ further (precisely, stability turns on the sign of $d\tilde L^2/dr$). Differentiating Step 3's result, $$\frac{d\tilde L^2}{dr} = \frac{2c^2 r_s\,r(r-3r_s)}{(2r-3r_s)^2}$$ which vanishes at $r=0$ (excluded), $r=3r_s$, and nowhere else for $r>3r_s/2$. This single interior root is the marginally stable radius: $$r_{\text{ISCO}} = 3r_s = \dfrac{6GM}{c^2}$$ For $r>3r_s$ the orbit is stable ($d\tilde L^2/dr>0$, larger $r$ needs more angular momentum, as in Kepler); for $3r_s/2<r<3r_s$ it is unstable.

Step 5 (the photon sphere from the null geodesic). For null geodesics the same substitution with $\kappa=0$ gives instead $$\left(\frac{dr}{d\lambda}\right)^2 = E^2 - \frac{L^2}{r^2}\left(1-\frac{r_s}{r}\right) \equiv E^2 - V^2_{\gamma}(r)$$ with $$V^2_\gamma(r) = L^2\left(\frac{1}{r^2}-\frac{r_s}{r^3}\right)$$ Setting the derivative of the bracket to zero, $$\frac{d}{dr}\left(\frac{1}{r^2}-\frac{r_s}{r^3}\right) = -\frac{2}{r^3}+\frac{3r_s}{r^4}=0 \;\Longrightarrow\; r=\frac{3r_s}{2}$$ directly locates the unique radius of circular photon orbits, $$r_{\text{ph}} = \dfrac{3}{2}r_s = \dfrac{3GM}{c^2}$$ matching the limit found in Step 3, and it is unstable because the bracket $1/r^2-r_s/r^3$ has a local maximum, not minimum, there — any inward nudge sends the photon spiraling into the horizon, any outward nudge sends it escaping to infinity.