Define an isotropic fourth-tensor in terms of Kronecker delta symbols
$$\eqalign{
\def\E{{\cal E}}
\def\G{{\cal S}} \def\H{{\cal D}} \def\T{{\cal T}}
\def\A{{\cal A}} \def\B{{\cal B}}
\def\d{\delta}
\def\LR#1{\left(#1\right)}
\def\qiq{\quad\implies\quad}
\def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}}
\E_{ikjl} &= \d_{ij}\,\d_{kl} \\
}$$
When doubly contracted with an arbitrary matrix $\,(A)\,$ it acts as the identity tensor
$$\eqalign{
A:\E &= \E:A = A \\
}$$
When singly contracted it exhibits useful rearrangement properties
$$\eqalign{
A\cdot B\cdot C
\;=\; \LR{A\cdot\E\cdot C^T}:B
\;=\; B:\LR{A^T\cdot\E\cdot C} \\
}$$
Applying this to your problem
$$\eqalign{
F &= A\cdot B \\
dF &= dA\cdot B \;=\; \LR{\E\cdot B^T}:dA \\
\grad FA &= {\E\cdot B^T} \\
}$$
Or in component form
$$\eqalign{
\grad{F_{ij}}{A_{kl}}
&= \sum_p \E_{ijkp} B^T_{pl} \\
&= \sum_p \d_{ik}\d_{jp}\,B_{lp} \\
&= \d_{ik}\,B_{lj} \\
}$$
In general, the double and single contraction products are defined as
$$\eqalign{
\H = \A:\B &\qiq \H_{ijmn} = \sum_k\sum_l \A_{ijkl}B_{klmn} \\
\G = \A\cdot\B &\qiq \G_{ijkmnp} = \sum_l \A_{ijkl}B_{lmnp} \\
}$$
The rightmost indices of the first tensor are summed with the leftmost indices of the second tensor, in the same order.
This idea can be extended to further contractions, e.g.
a triple contraction would be
$$\eqalign{
\T = \A\therefore\B \qiq \T_{im}
= \sum_j\sum_k\sum_l \A_{ijkl}B_{jklm} \\
}$$
