Coupled Ricci scalar action In the action where the Ricci curvature couples with the field $B^{\mu}$:
$$S=\int d^4x\sqrt{-g}\left(-\frac14F_{\mu\nu}F^{\mu\nu}+V(B_{\sigma}B^{\sigma}) +R\lambda B_{\mu}B^{\mu}\right)$$
To obtain the equations for the field $B^{\mu}$ I did the variation w.r.t. $B^{\mu}$ and got the following:
$$ 2B^\mu \left( \lambda R + \frac{\partial V(B^2)}{\partial B^2} \right)=F^{\mu\nu}_{\space\space\space\space,\nu}$$
Is there any problem with this approach given that $R$ is now coupled to the field $B^\mu$? Is the equation I obtained correct?
 A: To be clear, you can define any Lagrangian you want. Whether or not this is the right thing to do depends entirely on you and what you want.
With that in mind, the fact that $R$ is the Ricci scalar is immaterial to the question. The more interesting point to note is that no derivatives of your $B$ field appear and hence it represents a constraint on the system, like a Lagrange multiplier, rather than a new dynamical field.
But as I said, you can define whatever you want, and in particular you can define this.
Edit: Since it appears the OP intended $B$ to be the vector potential associated to the field strength $F$, my comments about $B$ representing a Lagrange multiplier are not applicable. Instead, my answer reduced only to "you can define whatever Lagrangian you like, whether this is a good idea or not is entirely dependent on the context of your specific problem."
A: The equation seems correct. The "hardest" part is to realise that:
$$\delta (V(B_{\sigma}B^{\sigma})) = \cfrac{\partial{V}}{\partial{B^2}} \delta(B_{\sigma}B^{\sigma})$$
and the rest is a standard calculation. Regarding the Ricci scalar part, when you vary with respect to a field the variation of all other fields is zero. Since this term is a scalar quantity you can have it in your lagrangian.
