Relationship between a (2,0)-type tensor and the differential of $f\in\mathcal{F}(M)$ Let $g$ be a (2,0)-type tensor in $M=\mathbb{R}^n$ such that, for each $p\in M$, the bilinear aplication $g:T_p M\times T_p M\rightarrow \mathbb{R}$ is non-degenerate. 
I must show that given $f\in\mathcal{F}(M)$, there exists a unique vector field $X^f\in\Xi(M)$ such that $df=g(X^f,\cdot)$, but I really don't even know how to start.
 A: The smooth function $f$ defines a one-form $df$ on $M$. Let $[g_{ij}]=A$ and suppose $A^{-1} = [g^{ij}]$. We can find such $A$ and $A^{-1}$ at each point given that $g$ is non-degenerate. Here I assume the existence of some coordinate system with which we can construct coordinate basis $\partial/\partial x^i |_p$ for each $p \in U$. Technically, you'll have to glue together my local result once we're done... In any event, $df = \sum \alpha_i dx^i$ where $\alpha_i = df(\partial_i)$. Let $X^f = \sum_{i,k} g^{ik}\alpha_k \partial_i$ at each $p \in U$,
$$ g(X^f, \partial_j) = \sum_{i,k} g^{ik}\alpha_k g(\partial_i, \partial_j) = 
\sum_{k} \left(\sum_i g^{ik}g_{ij}\right) \alpha_k$$
using bilinearity of $g$ and the definition $g(\partial_i,\partial_j) = g_{ij}$. By the definition of inverse matrices,
$$ g(X^f, \partial_j) = \sum_{i,k} \delta_{jk} \alpha_k = \alpha_j = df( \partial_j) $$
Hence, $g(X^f, \cdot ) = df$. If one coordinate system does not cover $M$ then you'll need to think about how to glue together $X^f$ as I construct on overlaps of coordinate systems. I leave that and other picky details to you.
