Relation involving the inverse of a symplectic form I am working on Exercise 14.34 on Penrose's "The Road to Reality". I need to prove the following result as a means to show Jacobi's identity for Poisson brackets (which I know how to do). In what follows I am using abstract index notation and $\nabla$ is some (arbitrary) torsion-free connection.
Let $S_{ab}$ be a $(0,2)$ non-singular antisymmetric tensor ($S_{ba}=-S_{ab}$) that satisfies $d$S$=0$ or equivalently $\nabla_aS_{bc} + \nabla_bS_{ca} + \nabla_cS_{ab} = 0$. Let also $S^{ab}$ be the inverse of $S_{ab}$ (namely, $S^{ab}S_{bc} = \delta^a_c$, where $\delta$ is Kronecker's delta tensor). I need to show the following:

$$S^{ab} \nabla_a S^{cd} + S^{ac} \nabla_a S^{db} + S^{ad} \nabla_a S^{bc} = 0$$

My understanding - although Penrose is not always clear about those things - is that I can show the result using just the assumptions above. In particular, without any use of a metric for index lowering. Is that possible?
What I tried was to differentiate $S^{ab}S_{bc} = \delta^a_c$ using Leibniz rule (aka the product rule) but this led me to a dead end as there is always lower index terms that I cannot get rid of.
Note: I am not necessarily looking for a full answer, as I want to figure out this myself. I am interested in a hint and also potentially making sure I am not missing any assumptions. Penrose is notoriously vague about those things sometimes (although this is otherwise some excellent and concise work).
 A: Differentiate $S^{ab}S_{bc} = \delta^a_c$ gives (I hope I did not mess up the sign)
$$ \nabla_c S^{ab} =  S^{ad} S^{be} \nabla_c S_{de}.$$
With that, you should be able to show
\begin{align}
S^{ab}& \nabla_a S^{cd} + S^{ac} \nabla_a S^{db} + S^{ad} \nabla_a S^{bc} \\
&= S^{ab} S^{ce}S^{df} \nabla_a S_{ef} + S^{ac} S^{de} S^{bf} \nabla_a S_{ef} + S^{ad} S^{be} S^{cf}\nabla_a S_{ef}
\end{align}
all the repeated indices (a, e, f) are summed over. Of course we want to use
$$ \nabla_a S_{ef} + \nabla_e S_{fa} + \nabla_f S_{ae} = 0,$$
and to do that, you can rename a, e, f on the second and third terms: e.g. renaming $a\to e$, $e\to f$, $f\to a$ respectively, we have
$$S^{ac} S^{de} S^{bf} \nabla_a S_{ef} = S^{\color{red}{e}c} S^{d\color{green}{f}} S^{b \color{pink}{a}} \nabla_{\color{red}{e}} S_{\color{green}{f}\color{pink}{a}}= S^{ab} S^{ce} S^{df} \nabla_{e} S_{fa},$$
where in the last equality we used $S^{ab}  = - S^{ba}$. One can similarly deal with the third terms and obtain
$$S^{ad} S^{be} S^{cf}\nabla_a S_{ef}= S^{ab} S^{ce} S^{df} \nabla_{f} S_{ae}.$$
Thus we have
$$S^{ab} \nabla_a S^{cd} + S^{ac} \nabla_a S^{db} + S^{ad} \nabla_a S^{bc} = S^{ad} S^{be} S^{cf} (\nabla_a S_{ef} + \nabla_e S_{fa} + \nabla_f S_{ae})=0.$$
