Ricci curvature: step in proof of a paper by Hamilton In Hamilton's paper "The Ricci Curvature Equation" (in Seminar on Nonlinear Partial Differential Equations, here), I can do all of Lemma 4.2 except for the following relation:
$$
-g^{ik}g^{j\ell}h_{pk}\partial_jF^p_{i\ell}=-\frac{1}{2}\Delta E+g^{ik}g^{j\ell}(\partial_jh_{kq})F^q_{i\ell}
$$
where $g$ and $h$ are metrics, $\Delta$ is the Laplacian w.r.t. $g$, and $E=g^{ij}h_{ij}$. 
From what I understand the relation follows from directly calculating the LHS:
$$
-g^{ik}g^{j\ell}h_{pk}\partial_jF^p_{i\ell}
$$
where
$$
F^p_{i\ell}=\frac{1}{2}h^{sp}(\partial_ih_{\ell s}+\partial_\ell h_{is}-\partial_sh_{i\ell}),
$$
however what I'm currently trying isn't working.
Edit: A correct answer has been given.
 A: This $\partial_i$ is used by Hamilton to denote the Levi-Civita connection with respect to the metric $g$, or with respect to $\Gamma_{i}{}^{k}_{j}$ 
Notice that there is a typo in the original paper in the fifth line on p.52 where it is said that "$\partial_i$ is covariant differentiation with respect to $F_{i}{}^{k}_{j}$".
Let me change the notation to make the things clearer. I use $\nabla^{(g)}$ and $\nabla^{(h)}$ for the Levi-Civita connections of the metrics $g$ and $h$ respectively. 
In other words, I use $\nabla^{(g)}_i$ instead of Hamilton's $\partial_i$ to avoid a possible confusion with the partial derivatives.
The difference of the connections $F_{i}{}^{k}_{j}$ in then expressed by the identity:
$$
\nabla^{(g)}_i \omega_j = \nabla^{(h)}_i \omega_j + F_{i}{}^{k}_{j} \omega_k
$$
where $\omega_i$ is an arbitrary $1$-form (covector).
Now we can calculate the Laplacian w.r.t. $g$ of the function $E = g^{i j} h_{i j}$. 
$$
\begin{align}
\Delta^{(g)} E & = g^{i j} \nabla^{(g)}_i \nabla^{(g)}_j (g^{k l} h_{k l}) \\
& = g^{i j} g^{k l} \nabla^{(g)}_i \nabla^{(g)}_j h_{k l} \\
& = g^{i j} g^{k l} \nabla^{(g)}_i \Big( \nabla^{(h)}_j h_{k l} + F_{j}{}^{p}_{k} h_{p l} + F_{j}{}^{p}_{l} h_{k p} \Big) \
\end{align}
$$
Now we observe that $\nabla^{(h)}_j h_{k l} = 0$ and recover the identity 
$$
\nabla^{(g)}_j h_{k l} = F_{j}{}^{p}_{k} h_{p l} + F_{j}{}^{p}_{l} h_{k p}
$$
Continuing this process we obtain
$$
\begin{align}
\Delta^{(g)} E & =  g^{i j} g^{k l} \nabla^{(g)}_i \Big( F_{j}{}^{p}_{k} h_{p l} + F_{j}{}^{p}_{l} h_{k p} \Big) \\
& =  g^{i j} g^{k l}  \Big( h_{p l} \nabla^{(g)}_i F_{j}{}^{p}_{k}  + F_{j}{}^{p}_{k} \nabla^{(g)}_i h_{p l} + h_{k p} \nabla^{(g)}_i F_{j}{}^{p}_{l}  + F_{j}{}^{p}_{l} \nabla^{(g)}_i h_{k p} \Big) \\
& = 2 \, g^{i j} g^{k l}   h_{p l} \nabla^{(g)}_i F_{j}{}^{p}_{k}  + 2\,  g^{i j} g^{k l} F_{j}{}^{p}_{k} \nabla^{(g)}_i h_{p l} 
\end{align}
$$
which is equivalent to the equation in the question.
This also confirms the identity in @Avitus's answer:
$$
\begin{align}
\Delta^{(g)} E & =  2 \, g^{i j} g^{k l}  \nabla^{(g)}_i \Big(   h_{p l} F_{j}{}^{p}_{k} \Big)
\end{align}
$$
A: Hint: considering the r.h.s. of the  equation in the OP we can write
$$g^{ik}g^{jl}(\partial_j h_{kq})F^{q}_{il}=g^{ik}g^{jl}\partial_j\left(
h_{kq}F^{q}_{il}\right)-g^{ik}g^{jl}h_{kq}(\partial_j F^{q}_{il});
$$
but 
$$-g^{ik}g^{jl}h_{kq}(\partial_j F^{q}_{il})=-g^{ik}g^{jl}h_{qk}(\partial_j F^{q}_{il})=-g^{ik}g^{jl}h_{pk}(\partial_j F^{p}_{il}),$$
i.e. the l.h.s. of the given equation in the OP, by symmetry of the tensor $h$ and changing names to repeated indices.
It remains to prove that
$$g^{ik}g^{jl}\partial_j\left(
h_{kq}F^{q}_{il}\right)=\frac{1}{2}\Delta(E), $$
by using the definition of the Laplace-Beltrami operator $\Delta$ w.r.t. $g$.
