The <Euler-Lagrange field equation> obtained by varying the <nonminimally coupled scalar field> is
$$
\boxed{\Box\Phi-2\xi R\Phi=0.}
$$
One can check <stress-energy conservation> directly. The divergence of the kinetic contribution is $(\Box\Phi)\nabla_b\Phi$, because second derivatives of a scalar commute. With $f=\Phi^2$, the <contracted Bianchi identity> gives $\nabla^aG_{ab}=0$, while the <Ricci identity> gives
$$
\Box\nabla_bf=\nabla_b\Box f+R_b{}^c\nabla_cf.
$$
Consequently
$$
\begin{aligned}
\nabla^a\left[fG_{ab}+g_{ab}\Box f-\nabla_a\nabla_bf\right]
&=G_{ab}\nabla^af+\nabla_b\Box f-\Box\nabla_bf\\
&=(G_{ab}-R_{ab})\nabla^af=-\frac12R\nabla_bf.
\end{aligned}
$$
Since $\nabla_bf=2\Phi\nabla_b\Phi$, the full <stress-energy tensor> satisfies
$$
\boxed{\nabla^aT_{ab}=(\Box\Phi-2\xi R\Phi)\nabla_b\Phi=0\quad\text{on shell}.}
$$

The general reason is the <diffeomorphism Noether identity for a scalar field>. Under the <Lie derivative> along a compactly supported <vector field> $X$, $\delta g_{ab}=2\nabla_{(a}X_{b)}$ and $\delta\Phi=X^a\nabla_a\Phi$. Invariance of the action under <diffeomorphisms> gives, after <integration by parts>,
$$
0=\int\sqrt{-g}\,X^b\left[-\nabla^aT_{ab}
+\frac1{\sqrt{-g}}\frac{\delta S}{\delta\Phi}\nabla_b\Phi\right]d^4x.
$$
Arbitrariness of $X$ gives the same divergence identity. Conservation requires the scalar equation of motion, with no need to impose the <Einstein field equations> for the background metric.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 309 / 4 / b / ii / Solution

A nonminimal scalar coupling includes an interaction between the field and spacetime curvature, such as $-\xi R\Phi^2$ in the <Lagrangian density>. The <Euler-Lagrange field equation> for $\mathcal L=-\frac12(\nabla\Phi)^2-\xi R\Phi^2$ is $\Box\Phi-2\xi R\Phi=0$. The coefficient is convention dependent: an action written with $-\frac12\xi R\Phi^2$ has $\xi$ instead of $2\xi$.

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