If $A_{ij}$ is skew-symmetric prove that $(A_{23}/\sqrt{g}, A_{31}/\sqrt{g}, A_{12}/\sqrt{g})$ are components of a contravariant vector This question follows from the exercises in the book Tensor Calculus A Concise Course by Barry Spain. I attempted the question many times, yet still unsure of how to proceed. Note that g is the determinant of the metric tensor. 
I tried writing the components of the vector ( which is to be proved to be contravariant) after a change of basis and noticed that upon expanding the determinant of the transformed metric tensor the Jacobian of transformation appear on the numerator. I tried expressing this determinant with Levi-Civita symbol in tensor notation to exploit the skew-symmetry of the tensor, yet failed and got stuck. The original excerpt from the book follow below:

 A: *

*The book apparently does not distinguish between tensors and pseudotensors, which becomes important when one includes coordinate transformations where the determinant of the Jacobian matrix $\frac{\partial x^{\prime j}}{\partial x^i}$ is negative. Also note that terminology, notation and conventions differ from author to author.  

*Given a metric (0,2) tensor
$$ \mathbb{g} ~=~\sum_{j,k=1}^ng_{jk}~\mathrm{d}x^j\odot\mathrm{d}x^k,\tag{1}$$
let 
$$g~:=~\det (g_{jk})_{1\leq i,j \leq n}\tag{2}$$ 
be the determinant. The square root $\sqrt{|g|}$ transforms as a density $$\sqrt{|g^{\prime}|}~=~\frac{\sqrt{|g|}}{|J|}, \qquad J~:=~\det(\frac{\partial x^{\prime j}}{\partial x^i})_{1\leq i,j \leq n}, \tag{3}$$
under a coordinate transformation. Obviously we could consider densities of other weights. Since $\sqrt{|g|}$ is one of the most basic densities, we declare for our purposes that it transforms with weight $+1$.

*Define a pseudovolume form
$$\Omega ~:=~\sqrt{|g|}\mathrm{d}x^1\wedge \ldots \wedge\mathrm{d}x^n~=~\frac{\sqrt{|g|}}{n!}\sum_{i_1, \ldots, i_n=1}^n \varepsilon_{i_1, \ldots, i_n} \mathrm{d}x^{i_1}\wedge \ldots \wedge\mathrm{d}x^{i_n},\tag{4}$$
where $\varepsilon_{i_1, \ldots, i_n}\in\{-1,0,1\}$ is the Levi-Civita permutation symbol. The symbol $\varepsilon^{i_1, \ldots, i_n}\in\{-1,0,1\}$ with upper indices is the same as $\varepsilon_{i_1, \ldots, i_n}\in\{-1,0,1\}$ with lower indices, modulo overall sign conventions. It is straightforward to check that under a coordinate transformation, the pseudovolume form (4) transforms as
$$\Omega^{\prime} ~=~{\rm sgn }(J)~\Omega. \tag{5}$$

*It follows that the Levi-Civita permutation symbol $\varepsilon_{i_1, \ldots, i_n}$ is a pseudotensor density of weight $-1$, while 
$$ {\cal E}_{i_1, \ldots, i_n}~:=~ \sqrt{|g|}\varepsilon_{i_1, \ldots, i_n}\tag{6} $$
is a $(0,n)$ pseudotensor. Similarly,
$$ {\cal E}^{i_1, \ldots, i_n}~:=~ \frac{1}{\sqrt{|g|}}\varepsilon^{i_1, \ldots, i_n}\tag{7}$$
is a $(n,0)$ pseudotensor.

*Let us now return to OP's question in $n=3$ dimensions.
Given a 2-form 
$$\mathbb{A}~=~\frac{1}{2}\sum_{j,k=1}^3A_{jk}~\mathrm{d}x^j\wedge\mathrm{d}x^k,\tag{8}$$
i.e. a $(0,2)$ tensor. 
We can construct a pseudovector, i.e. a $(1,0)$ pseudotensor,
$$A^i~:=~\frac{1}{2}\sum_{j,k=1}^3{\cal E}^{ijk}A_{jk}~=~\frac{1}{2\sqrt{|g|}}\sum_{j,k=1}^3 \varepsilon^{ijk}A_{jk}, \tag{9} $$ 
essentially using the Hodge-star.
It transforms as
$$A^{\prime j}~=~{\rm sgn }(J)\sum_{i=1}^3 \frac{\partial x^{\prime j}}{\partial x^i}A^i\tag{10}$$ 
under a coordinate transformation.  
