= Solution
For a homogeneous <scalar field>, every spatial derivative vanishes. The <stress-energy tensor> consequently has no <momentum density> or directional stress:
$$
T^{0i}=0,\qquad T^{ij}=Pa^{-2}\delta^{ij},\qquad
\rho=\frac12\dot\phi^2+V,\quad P=\frac12\dot\phi^2-V.
$$
The only spatial tensor left is $\delta^{ij}$, so the stress is isotropic. This is the <perfect fluid> form in its comoving frame. Now expand the time component of <stress-energy conservation> using the <covariant derivative>:
$$
0=\nabla_\mu T^{\mu0}
=\partial_\mu T^{\mu0}+\Gamma^\mu{}_{\mu\lambda}T^{\lambda0}
+\Gamma^0{}_{\mu\lambda}T^{\mu\lambda}.
$$
The first term is $\dot\rho$. In the second, only $\lambda=0$ contributes, and $\sum_i\Gamma^i{}_{i0}=3H$, giving $3H\rho$. The third is
$$
\sum_{i,j}(a\dot a\delta_{ij})(Pa^{-2}\delta^{ij})=3HP.
$$
Here the quoted <Christoffel symbols> include their lower-index symmetry $\Gamma^i{}_{j0}=\Gamma^i{}_{0j}$. Therefore the <cosmological continuity equation> is
$$
\boxed{\dot\rho=-3\frac{\dot a}{a}(\rho+P).}
$$
As a check using the <scalar field> equation, $\dot\rho=\dot\phi(\ddot\phi+V_{,\phi})=-3H\dot\phi^2=-3H(\rho+P)$, with exactly the same sign.
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