Let $M$ be a smooth manifold and $f$, $g$ be smooth functions in some neibourhood of a point $x_0\in M$, $\nabla g\ne0$.
1) How to define $\displaystyle \frac{\partial f}{\partial g}$ invariantly? If $M$ is a domaqin in $\mathbb R^n$ then the derivative in the direction of $\nabla g$ seems to give an answer: $\displaystyle \frac{(\nabla f,\nabla g)}{|\nabla g|^2}$. But to calculate $\nabla g$ and $|\nabla g|$ on a manifold one needs a metric. From the other hand, if we consider smooth coordinates $(g_1=g, g_2,\ldots,g_n)$ in some neibourhood of $x_0$, then partial derivatives $\displaystyle \frac{\partial f}{\partial g}$ seem to be defined in the standard way. But the question arises, would the value $\displaystyle \frac{\partial f}{\partial g}$ be independent from the choice of $ g_2,\ldots,g_n$? If no, what are the correct way do do it? Is there some reference?
2) Let $f_1, f_2,\ldots,f_n$ be a smooth coordinates in some neibourhood of $x_0$. What is the object $(\displaystyle \frac{\partial f_1}{\partial g},\ldots,\frac{\partial f_n}{\partial g})$? Would it be by some chance a section of some good fiber bundle? Is there some reference where such objects are considered?
Thanks in advance!
Addition of may 27
Suppose that now there is a riemannian metric on $M$. Then what would be most natural definition of partial derivative $\displaystyle \frac{\partial f}{\partial g}$? For expample, to take $\displaystyle \frac{\partial f}{\partial g}=df(\nabla g)=(df,*dg)$ seems not right since that would mean $\displaystyle \frac{\partial f}{\partial g}=\displaystyle \frac{\partial g}{\partial f}$. If we take $\displaystyle \frac{(\nabla f,\nabla g)}{|\nabla g|}$ it would be sort of directional derivative. So the right question seems to me here is what value of $a$ it is best to take for $\displaystyle \frac{(\nabla f,\nabla g)}{|\nabla g|^a}$ to be a partial derivative? Or is there no the "best" choice? I tried to apply to dimensional analysis and it seems $a=1$ is the choice to have the result be like $\nabla f$ but I'm not sure because may be one have to assign some dimension to metric.