Question: Let $(M,g)$ be a Riemannian manifold and $g$ the Riemannian metric in coordinates $g=g\_{\alpha \beta}dx^{\alpha} \otimes dx^{\beta}$, where $x^{i}$ are local coordinates on $M$. Denote by $g^{\alpha \beta}$ the inverse components of the inverse metric $g^{-1}$. Let $\nabla$ be the Levi-Civita connection of the metric $g$. Consider, locally, the function $\det((g\_{\alpha \beta})\_{\alpha \beta})$. It is known that $\nabla \det((g\_{\alpha \beta})\_{\alpha \beta}) = 0$ by using normal coordinates etc...

I would like to show this fact without using normal coordinates. Just by computation. Here is what I have so far: 

$$\nabla \det((g\_{\alpha \beta})\_{\alpha \beta}) = \left [ g^{\gamma \delta} \partial\_{\delta} \det((g\_{\alpha \beta})\_{\alpha \beta}) \right ] \partial\_{\gamma} = \left [ \det((g\_{\alpha \beta})\_{\alpha \beta}) g^{\gamma \delta} g^{\beta \alpha} \partial\_{\delta} g\_{\alpha \beta}\right ] \partial\_{\gamma}.$$

Here: the first equality sign follows from the definition of the gradient of a function and the second equality sign is the derivative of the determinant.

**Question:** How do I continue from here without using normal coordinates? Or are there any mistakes? If yes, where and which?

Answer: 
Better to specify your direction vector with $\theta$ and $\varphi$ in [spherical coordinates](https://en.wikipedia.org/wiki/Spherical_coordinate_system) $(r,\theta,\varphi)$. Otherwise you have a problem with how to specify which of the three coordinates is omitted $-$ this requires another parameter.

But in fact a line only has four degrees of freedom, not five. This is because you can choose any point on the line to represent it. Actually finding four numbers to represent a line is a bit messy; start with the direction $(\theta,\varphi)$, and consider the plane $L$ through the origin that is perpendicular to it. Then the line is specified by its point of intersection with $L$. This requires only two coordinates, but you need some linear algebra to set up your coordinate system on $L$.

Perhaps somebody knows a more elegant parametrisation?
The idea is right, but, in general, we cannot fix a given coordinate of the direction vector to be $0$ or $1$. So usually we assume that the direction vector $(d\_1,d\_2,d\_3)$ has unit length, that means that $d\_1^2+d^2\_2+d^2\_3=1$.