curl(fF) with Einstein Summation Notation I considered the $k$th component of $\text{curl $f\mathbf{F}$}$. $f$ is a scalar field and $\mathbf{F}$ a vector field. 
$\color{green}{[}\nabla \times (fF)\color{green}{]} _{\LARGE{\color{green}{k}}} = \epsilon_{ij\LARGE{\color{green}{k}}}\partial_i(f\mathbf{F})_j $
$= \epsilon_{ij\LARGE{\color{green}{k}}}\partial_i(fF_j) \qquad \qquad \qquad \qquad (\text{since $\mathbf{(F)}_j :=$ the $j$th component of $\mathbf{F} = F_j$})$
$= \epsilon_{ij\LARGE{\color{green}{k}}}(F_j\partial_if + f\partial_iF_j) \qquad \qquad  (\text{since $f$ scalar})$
$= \underbrace{\epsilon_{ij\LARGE{\color{green}{k}}}F_j\partial_if}_{\Large{\bigstar}} + f\underbrace{{\epsilon_{ij\LARGE{\color{green}{k}}}\partial_iF_j}}_{\LARGE{\color{green}{[}\nabla \times \mathbf{F}\color{green}{]}_{\LARGE{\color{green}{k}}}}} $  
Hereafter, I refer only to the term with the star underneath.  Since [$\color{#007FFF}{F_j}$ corresponding to $\color{#007FFF}{\mathbf{F}}$] appears before [$\color{#FF00FF}{\partial_if}$ corresponding to $\color{#FF00FF}{\nabla f}$], thus ${\epsilon_{ij\LARGE{\color{green}{k}}}\color{#007FFF}{F_j}\color{#FF00FF}{\partial_if}} = {\color{green}{[}\color{#007FFF}{\mathbf{F}} \times \color{#FF00FF}{\nabla f}\color{green}{]} _{\LARGE{\color{green}{k}}}}$.
But the answer states $\color{green}{[}\nabla f \times \mathbf{F}\color{green}{]} _{\LARGE{\color{green}{k}}}$. What went wrong?

$\large{\text{Supplement to Andrew D's response :}}$
Here's my understanding of your answer : In ${\epsilon_{ij\LARGE{\color{green}{k}}}F_j\partial_if}, \; {(i, j, \LARGE{\color{green}{k}})}$ (in the subscript of the Levi-Civita symbol) denotes the order of the components. So the $i$th component must appear first, and the $j$th component second.
However, since $(i, j, k) = \color{brown}{(j, k, i)}$, therefore  $\epsilon_{ijk} = \epsilon_{\color{brown}{\LARGE{jki}}}$. $\color{brown}{\text{Now, $j$ precedes $i$, so wouldn't this result in the wrong order of the components?}}$

$\large{\text{2nd Supplement to Andrew D's Comment beneath his Answer :}}$
$\color{#3EB489}{\text{The variable in the permutation succeeding the variable that's not summed}}$ corresponds to the first component to appear. Here, $k$ denotes the component being analysed so is not summed. Since I am looking at $\color{brown}{(j, k, i)}$, $\color{#3EB489}{i}$ succeeds $k$ so the $\color{#3EB489}{i}$th component is the first. Therefore, ${\epsilon_{ijk}F_j\color{#3EB489}{\partial_{\LARGE{i}}}f}$ = ${\epsilon_{\color{brown}{\LARGE{jki}}}\color{#3EB489}{\partial_{\LARGE{i}}}F_jf} = \color{green}{[}\color{#3EB489}{\nabla f} \times \mathbf{F}\color{green}{]} _{\LARGE{\color{green}{k}}} $
However, this appears to discord with Steven Stadnicki's 2nd comment, according to which: $ {\epsilon_{\color{brown}{\LARGE{jki}}}F_j\partial_if} = {\color{green}{[}\mathbf{F} \times \nabla f\color{green}{]} _{\LARGE{\color{green}{k}}}}$?
 A: It isn't the ordering of $F_j$ and $\partial_if$ in the product of terms which determines what order the terms are in the cross product, it is the ordering of the suffix's which does so - as we have $\{i,j,k\}$ as our right-handed set (taken from the ordering of the suffices from the Levi-Civita symbol given), it means that we take the $\partial_if$ as being the first component of the cross product and $F_j$ as our second, so we get $[ \nabla f \times \mathbf F]_k$ as required.
If this doesn't make much sense, say so and I'll try and clarify what I'm saying.
A: Terms as $\epsilon_{ijk}a_ib_i$ do not carry any information on operations between $a_i$ and $b_j$, hence no information on e.g. commutativity. They really are meant as summing symbols (maybe with no meaning at al)l $F_j$ over indices which appear twice, no matter in what order these symbols are arranged:
$$\epsilon_{ijk}a_ib_j=a_i\epsilon_{ijk}b_j=b_ja_i\epsilon_{ijk}=\dots$$
(Maybe, writing those kind of sums explicitly for e.g. $i,j,k=1,2,3$ helps.)
Concerning your 2nd supplement, you only need to keep in mind that $$(\nabla f)_i:=\partial_if:=\frac{\partial f}{\partial x^i}$$
denotes one single symbol with index. Therefore, $\epsilon_{ijk}\partial_iF_jf$ does not make sense in this context. 
Finally, Steven Stadnicki's first comment is most important.
A: Recall that $\vec{a} \times \vec{b} = - \, \vec{b} \times \vec{a}$. 
Now consider each, using your definitions (I have included the unit vector $\hat{e}_i$ explicitly, simply for additional clarity; see below):
$$ \vec{a} \times \vec{b} = \epsilon_{i{\color{red}j}{\color{blue}k}} \, \hat{e}_i \, a_{\color{red}j} \, b_{\color{blue}k}
= - \, \vec{b} \times \vec{a} = - \epsilon_{i{\color{red}j}{\color{blue}k}} \, \hat{e}_i \,  b_{\color{red}j} \, a_{\color{blue}k}.$$
Recall that $\epsilon_{ijk}$ is perfectly anti-symmetric, which means that interchanging any two indices changes its sign (there are only 6 nonzero terms, so it's not long to verify this by plugging in actual values), for example: 
$$\epsilon_{1{\color{red}2}3} = - \epsilon_{13{\color{red}2}}.$$
Or, in general, $-\epsilon_{i{\color{red}j}k} = \epsilon_{ik{\color{red}j}}$. Put this together with our result for $-\vec{b} \times \vec{a}$, above, we have: 
$$- \, \vec{b} \times \vec{a} = - \epsilon_{i{\color{red}j}{\color{blue}k}} \, \hat{e}_i \,  b_{\color{red}j} \, a_{\color{blue}k} = \epsilon_{i{\color{blue}k}{\color{red}j}}\, \hat{e}_i \,a_{\color{blue}k} \, b_{\color{red}j} = \vec{a} \times \vec{b},$$
as expected.
N.B. the ordering of the $a_{j/k}$ and $b_{j/k}$ terms themselves doesn't matter, it's the relative position of their index in $\epsilon_{ijk}$ that matters. ie.
$$\begin{align*}\vec{a} \times \vec{b} & = \epsilon_{ijk} \, \hat{e}_i \, a_j \, b_k \\
& = \epsilon_{ijk} \hat{e}_i \, b_k \, a_j \\
& = \epsilon_{jki} \hat{e}_j \, a_k \, b_i \\
& = \epsilon_{kij} \hat{e}_k \, a_i \, b_j
\end{align*}
$$
In the beginning, it can be easy to conflate $i$, $j$ and $k$ with the unit vector in Cartesian space, but they aren't the same thing. 
Notice this means you can cycle through the equivalent forms of epsilon until you find a form that "matches" the cross product. So, in your case: 
$$\begin{align*}
& \quad \epsilon_{ijk} F_j \partial_i f \\
& = \epsilon_{jki} F_j\partial_i f\\ 
& = \epsilon_{kij} F_j\partial_i f
\end{align*}$$
Which we can now read as $\nabla f \times \vec{F}$ (or, equivalently: $-\, \vec{F} \times \nabla f$).  

On including the unit vector in the definition of the cross product (because it seems there is some conflation between the indices $i$, $j$, $k$ and the basis vectors $\hat{i} = \hat{e}_1$, $\hat{j} = \hat{e}_2$ and $\hat{k}=\hat{e}_3$).
I chose to include the unit vector in my definition, because it removes this "components" expression. Instead, we can just give an expression  for the (entire) cross product:
$$ \vec{a} \times \vec{b} = \epsilon_{ijk} \, \hat{e}_i \, a_j \, b_k.$$
This is the same as your definition; however, I've included the unit vector $\hat{e}_i$ explicitly. Notice, for example when $i = 1$, we get the first component of the cross-product: 
$$ \epsilon_{123} \hat{e}_1 a_2b_3 + \epsilon_{132} \hat{e}_1 a_3b_2
   = \hat{e}_1\, a_2b_3 - \hat{e}_1 \,a_3b_2 = (a_2b_3 - a_3b_2) \,\hat{e}_1,$$
and similarly for the other two components. 
However, the naming of the indices is totally arbitrary, so we could equally write: 
$$ \vec{a} \times \vec{b} = \epsilon_{lmn} \, \hat{e}_l \, a_m \, b_n, $$
or even,
$$ \vec{a} \times \vec{b} = \epsilon_{kji} \, \hat{e}_k \, a_j \, b_i, $$
because this represents every possible combination for all three indices equal to any one of 1, 2, or 3. (so $3^3 = 27$ different terms, where all but 6 are zero). 
Now, once you've chosen your indices, order does matter, in particular all cyclic permutations [(1,2,3), (2,3,1), (3,1,2)] will have the same value, and all acyclic [(1,3,2), (3,2,1) and (2,1,3)] permutations will have the same value, i.e. 
$$\vec{a} \times \vec{b} = \epsilon_{i{\color{blue}j}{\color{red}k}}\, \hat{e}_i \, a_{{\color{blue}j}} b_{\color{red}k} 
= \epsilon_{j{\color{blue}k}{\color{red}i}}\, \hat{e}_j \, a_{{\color{blue}k}}b_{\color{red}i} 
= \epsilon_{k{\color{blue}i}{\color{red}j}}\,\hat{e}_k \,  a_{\color{blue}i} b_{\color{red}j}$$
whereas 
$$ \epsilon_{i{\color{red}k}{\color{blue}j}}\,\hat{e}_i \,  a_{{\color{blue}j}} b_{\color{red}k} 
= \epsilon_{j{\color{red}i}{\color{blue}k}}\, \hat{e}_j \, a_{{\color{blue}k}}b_{\color{red}i} 
= \epsilon_{k{\color{red}j}{\color{blue}i}}\, \hat{e}_k \, a_{\color{blue}i} b_{\color{red}j} = \vec{b} \times \vec{a}$$
Hope this helps!
