Curl(curl(A)) with Einstein Summation Notation (subscript & superscript !) I am a rookie so I hope you know the answer!
Okay so from C. Möller and Landau & Lifshitz I have gathered that:
$$
\gamma_{ij}=g_{ij}-\frac{g_{0i}g_{0j}}{g_{00}} \quad \gamma^{ij}\gamma_{jk}=\delta^i_k\quad \gamma^{ij}=g^{ij}\quad\gamma = \textbf{det}(\gamma_{ab}) = \frac{1}{\textbf{det}(\gamma^{ab})}\\
\epsilon_{ijk} = \epsilon^{ijk}=\left\{\begin{array}{ll}
1 & \text{if } (i,j,k) \text{ is an even permutation of } (1,2,3) \\
-1 &\text{if } (i,j,k) \text{ is an odd permutation of } (1,2,3) \\
0 & \text{otherwise}
\end{array}\right.\\
\varepsilon_{ijk}=\sqrt{\gamma}\epsilon_{ijk}\quad \varepsilon^{ijk}=\frac{1}{\sqrt{\gamma}}\epsilon^{ijk}\\
[\textbf{curl}(\textbf{v})]^{i}=\varepsilon^{ijk} v_{k,j}=\frac{1}{2}\varepsilon^{ijk}(v_{k,j}-v_{j,k})\quad\textbf{div}(\textbf{v})=\frac{1}{\sqrt{\gamma}}[\sqrt{\gamma} v^i]_{,i}
$$
Can you similarly define
$$
[\mathbf{curl}(\mathbf{\tilde{v}})]_i =\varepsilon_{ijk}v^{k,j}
$$
And thereafter 
$$
[\mathbf{curl}\circ\mathbf{curl}(\mathbf{\tilde{v}})]^a=\varepsilon^{abc} [\varepsilon_{cde}v^{e,d}]_{,b}\\
[\mathbf{curl}\circ\mathbf{curl}(\mathbf{v})]_a=\varepsilon_{abc} [\varepsilon^{cde}v_{e,d}]^{,b}
$$
And if so, what is the partial derivative of $\sqrt{\gamma}$ and $\frac{1}{\sqrt{\gamma}}$? Is it valid to reexpress the equations above as 
$$
[\mathbf{curl}\circ\mathbf{curl}(\mathbf{\tilde{v}})]^a=\varepsilon^{abc} \varepsilon_{cde}v^{e,d}_{\quad,b}\\
[\mathbf{curl}\circ\mathbf{curl}(\mathbf{v})]_a=\varepsilon_{abc} \varepsilon^{cde}v_{e,d}^{\quad,b}
$$
or is that wrong? Realized a commaderivative upstairs is not defined. Would it be expressed with codifferential somehow?
(http://en.wikipedia.org/wiki/Ricci_calculus#Raised_and_lowered_indices for summation convention)
 A: I would answer to your question by collecting some facts on the structures you introduce above.
- A typo
The relations
$$\epsilon_{ijk}=\sqrt{\gamma}\epsilon_{ijk}, $$
$$\epsilon^{ijk}=\frac{1}{\sqrt{\gamma}}\epsilon^{ijk}, $$
are not true, unless $\sqrt{\gamma}=1$.
- On curl on $v=v_ie_i$.
The curl operator on vectors $v=v_ie_i$ in $\mathbb R^3$ gives the vector
$$\left(\operatorname{curl}(v)\right)_i=\epsilon_{ijk}\partial_jv_k; $$
such $\operatorname{curl}(v)$ is equal to your $\boldsymbol{\operatorname{curl}}(v)$ as $\epsilon_{ijk}=\epsilon^{ijk}$. Then
$$\boldsymbol{\operatorname{curl}}\left(\boldsymbol{\operatorname{curl}}(v)\right)_i=\operatorname{curl}(\left(\operatorname{curl}(v)\right)_i=\epsilon_{ijk}\partial_j
(\epsilon_{krs}\partial_rv_s)=
\epsilon_{ijk}\epsilon_{krs}\partial_j
\partial_rv_s; $$
using the relations 
$$\epsilon_{ijk}\epsilon_{krs}=\epsilon_{kij}\epsilon_{krs}=\delta_{ir}\delta_{js}-\delta_{is}\delta_{jr},$$
one arrives at
$$\operatorname{curl}(\left(\operatorname{curl}(v)\right)_i=
(\delta_{ir}\delta_{js}-\delta_{is}\delta_{jr})\partial_j
\partial_rv_s=\partial_s\partial_iv_s-\sum_r\partial^2_rv_i,  $$
or
$$\boldsymbol{\operatorname{curl}}(v)=\nabla(\operatorname{div}(v))-\nabla^2(v)$$
- On curl on $\tilde{v}=v^ie_i$.
The curl operator on vectors $v=v^ie_i$ in $\mathbb R^3$, where
$$v^i=g^{in}v_n, $$
is the vector
$$\left(\boldsymbol{\operatorname{curl}}(\tilde{v})\right)_i=\epsilon_{ijk}\partial_j(g^{kn}v_n); $$
then
$$\boldsymbol{\operatorname{curl}}\left(\boldsymbol{\operatorname{curl}}(\tilde{v})\right)_i=\epsilon_{ijk}\partial_j
(\epsilon_{krs}\partial_r(g^{rq}v_q))=
\epsilon_{ijk}\epsilon_{krs}\partial_j
(\partial_rg^{rq}v_q+g^{rq}\partial_rv_q). $$
Using once again the relations $$\epsilon_{ijk}\epsilon_{krs}=\delta_{ir}\delta_{js}-\delta_{is}\delta_{jr}$$ you can arrive at the result you are looking for. 
