Relationship between Jacobian and scale factors of an orthogonal system in tensor notation. Given a curvline coordinate system in $R^3$ with parameters $(q_1,q_2,q_3)$, I have to prove that the Jacobian of the transformation $J(\frac{x_1,x_2,x_3}{q_1,q_2,q_3})$ is equivalent to the product of the metric factors $h_1h_2h_3$, and I have to use tensor notation.
The metric factors of the system are given by: $h_{ij}^2=\frac{\partial x_k}{\partial q_i}\frac{\partial x_k}{\partial q_j}$. Since it's an orthogonal system, I can apply Kroneckers Delta $\delta_{ij}$ to get $\delta_{ij}h_{ij}^2=h_{i}^2=\frac{\partial x_k}{\partial q_i}\frac{\partial x_k}{\partial q_i}$.
On the other hand, the Jacobian matrix is given by $J_{lm}=\frac{\partial x_l}{\partial q_m}$, so the Jacobian would be given by $J=\epsilon_{lmn}\frac{\partial x_1}{\partial q_l}\frac{\partial x_2}{\partial q_m}\frac{\partial x_3}{\partial q_n}$.
I have to proof either $J=h_1h_2h_3$ or $J^2=h_1^2h_2^2h_3^3$, but i don't know how to do it.
Please help me! Thanks in advance!
 A: The tangent basis vectors of your system is given by the columns of the Jacobian matrix. With a little abuse of notation we can write
$$\mathbf{r}_1=J^m_1,\,\mathbf{r}_2=J^v_2,\,\mathbf{r}_3=J^w_3$$
Where $J^m_1$ is the first column of the Jacobian matrix and so on. For an orthogonal system we also have some additional properties:
$$\mathbf{r}_1\cdot \mathbf{r}_2=0,\,\mathbf{r}_1\cdot \mathbf{r}_3=0,\,\mathbf{r}_2\cdot \mathbf{r}_3=0$$
The scale factor  $h_\alpha$ for each basis vector $\mathbf{r}_\alpha$ is just
$$h_\alpha^2=||\mathbf{r}_\alpha||^2=g_{mu}J^m_\alpha J^u_\alpha$$
Notice that the greek letter $\alpha$ is not part of the summation convention in this specific context.
From basic linear algebra we remember that for a a right oriented base we must also have
$$\mathbf{r}_2\times \mathbf{r}_3=\frac{h_2h_3}{h_1}\mathbf{r}_1$$
In index notation this "cross product" can be interpreted as
$$(\mathbf{r}_2\times \mathbf{r}_3)_u=\varepsilon_{uvw}J^v_2J^w_3=\frac{h_2h_3}{h_1}J^m_1g_{mu}$$
Now we can see that the Jacobian $J$ is
$$J=\varepsilon_{uvw}J^u_1J^v_2J^w_3=J^u_1g_{mu}J^m_1\frac{h_2h_3}{h_1}=(h_1)^2\frac{h_2h_3}{h_1}=h_1h_2h_3$$
This is of course the familiar scalar triple product $J=\mathbf{r}_1\cdot(\mathbf{r}_2\times \mathbf{r}_3)$
Edit: As I mentioned in the comments, an alternative to this approach is to use the fact that the "metric factors" squared you mention in your question is actually just $g_{hk}=\sum_{j=1}^n\frac{\partial x^j}{\partial q^h}\frac{\partial x^j}{\partial q^k}$.
This in turn is just the transposed Jacobian matrix times itself ($J^TJ)$. And for an orthogonal system this is a diagonal matrix (since $g_{hk}=\mathbf{r}_h\cdot \mathbf{r}_k$).
Using some properties of determinants, we understand that $g=J^2$ since $\det(J^TJ)=\det(J)\det(J)$.
The determinant of a diagonal (or even an upper triangular) matrix is the product of its diagonal entries.
So what we are looking for the product of the diagonal elements of the matrix $g_{hk}$, ie the product $g_{11}g_{22}g_{33}$
$$J^2=(\mathbf{r}_1\cdot\mathbf{r}_1)(\mathbf{r}_2\cdot\mathbf{r}_2)(\mathbf{r}_3\cdot\mathbf{r}_3)=h_1^2h_2^2h_3^2$$
