Ricci tensor from this action

Here is an action for a theory which couples gravity to a field in this way:$$S = \int d^4 x \ \sqrt{-g} e^{\Phi} (R + g^{ab} \Phi_{;a} \Phi_{;b})$$I determine\begin{align*}
\frac{\partial L}{\partial \phi} &= \sqrt{-g} e^{\Phi} (R + g^{ab} \Phi_{;a} \Phi_{;b}) \\
\nabla_a \frac{\partial L}{\partial(\nabla_a \Phi)} &= 2\sqrt{-g} e^{\Phi} g^{ab} (\Phi_{;a} \Phi_{;b} + \Phi_{;ba})
\end{align*}giving ##R = g^{ab} (\Phi_{;a} \Phi_{;b} + 2\Phi_{;ba})##. Now vary the action with respect to the metric,\begin{align*}
\frac{\delta S}{\delta g^{ab}} &= -\frac{1}{2}\sqrt{-g} e^{\Phi} g_{ab} (R + g^{cd} \Phi_{;c} \Phi_{;d}) + \sqrt{-g} e^{\Phi}(\frac{\delta R}{\delta g^{ab}} + \Phi_{;a} \Phi_{;b}) \end{align*}Put ##\delta R_{ab} / \delta g^{ab} = R_{ab}## and insert the previous equation for ##R##. Zero the variation and cancel the common factor ##\sqrt{-g} e^{\Phi}##,$$0 = -g_{ab} g^{cd} (\Phi_{;c} \Phi_{;d} + \Phi_{;dc}) + R_{ab} + \Phi_{;a} \Phi_{;b}$$This should give ##R_{ab} = \Phi_{;ba}## but it doesn't work out that way because the indices are mangled. Can somebody see the error?

ergospherical said:

Here is an action for a theory

Is there a particular source from which you got this? It looks like it might be the general action for one kind of scalar-tensor theory of gravity.

It's just a very old exam. I presumed it was just something arbitrary the examiner thought up, but may have some application.