= Solution
Let $M_{\rm Pl}$ be the <reduced Planck mass>, $M_{\rm Pl}^{-2}=8\pi G$ in units $\hbar=c=1$. Combining that equation with the <Friedmann acceleration equation> gives
$$
\dot H=-\frac{\rho+P}{2M_{\rm Pl}^2}=-\frac{\dot\phi^2}{2M_{\rm Pl}^2}.
$$
Differentiate $3M_{\rm Pl}^2H^2=\rho$ to obtain $\dot\rho=-3H(\rho+P)=-3H\dot\phi^2$. For the homogeneous canonical <scalar field>, $\dot\rho=\dot\phi\ddot\phi+V_{,\phi}\dot\phi$, so
$$
\dot\phi\bigl(\ddot\phi+3H\dot\phi+V_{,\phi}\bigr)=0.
$$
Where $\dot\phi\ne0$, division gives the <inflaton equation of motion>; continuity extends it to isolated turning points:
$$
\boxed{\ddot\phi+3H\dot\phi=-V_{,\phi}.}
$$
The <Hubble friction> term describes loss of kinetic energy into expansion.
There is a small logical caveat if the field is frozen on an entire interval: the homogeneous Einstein constraints alone only give the product equation and do not force $V_{,\phi}=0$. The <frozen scalar fields and the Friedmann constraint> counterexample makes this explicit. For a canonical scalar, the field equation also follows directly from its <inflaton action in an expanding universe>, whose homogeneous Lagrangian is $a^3[\dot\phi^2/2-V(\phi)]$. The <Euler-Lagrange equation> gives $d(a^3\dot\phi)/dt+a^3V_{,\phi}=0$, including the frozen case, and yields the same boxed equation without division by $\dot\phi$.
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