Does $d x^\mu=g^{\mu\nu} \partial_\nu$, or $\partial_\nu=g_{\nu \mu}dx^\mu$? As the title suggests: does $d x^\mu=g^{\mu\nu} \partial_\nu$, or $\partial_\nu=g_{\nu \mu}dx^\mu$?
This is merely my own guess, from the material that I'm reading; so I'm not sure... The partial derivative operator is still an differential operator in my point of view.
This may be too basic for you, but thank you very much.
 A: Not quite. Although the index balance is correct, in ${\rm d}x^\mu = g^{\mu\nu}\partial_\nu$ you have that the left side is a $1$-form, while the right side is a vector field. What happens here is that if $(M,g)$ is your pseudo-Riemannian manifold (which I assume that in your case should just be a spacetime), the metric induces the so called musical isomorphisms $\flat\colon T_x M \to T_x^*M$ and $\sharp \colon T^*_xM \to T_xM$ for every point $x \in M$. Said isomorphisms are characterized by $v_\flat(w) = g_x(v,w)$ for $v, w \in T_xM$, and also for $\xi \in T_x^*M$, the vector $\xi^\sharp \in T_xM$ is characterized by the relation $\xi(v) = g_x(v, \xi^\sharp)$, for all $v \in T_xM$. You can put all of these isomorphisms together to get maps $\flat : TM \to T^*M$ and $\sharp\colon T^*M\to TM$ between tangent bundles (which I'm still denoting by the same symbols), and then they can be "upgraded" to the section level, yielding isomorphisms $\flat\colon \mathfrak{X}(M) \to \Omega^1(M)$ and $\sharp\colon \Omega^1(M) \to \mathfrak{X}(M)$ between vector fields and $1$-forms. What you actually have for a given coordinate system are the formulas $${\rm d}x^\mu = g^{\mu\nu} (\partial_\nu)_\flat \quad\mbox{and}\quad \partial_\mu = g_{\mu \nu}({\rm d}x^\nu)^\sharp.$$What happens is that people just sweep these isomorphisms under the rug and write just the formulas you have seen in your textbook. For example, to prove the first one, you just have to check that both sides act the same in an arbitrary coordinate vector field $\partial_\lambda$, say. Indeed, we have $$g^{\mu\nu}(\partial_\nu)_\flat(\partial_\lambda) = g^{\mu\nu}g(\partial_\nu,\partial_\lambda) = g^{\mu\nu}g_{\nu\lambda} = \delta^\mu_{\;\lambda} = {\rm d}x^\mu(\partial_\lambda),$$as wanted. The second one follows from the first one (up to index relabeling) by doing $${\rm d}x^\mu = g^{\mu\nu}(\partial_\nu)_\flat \implies g_{\lambda \mu}{\rm d}x^\mu = g_{\lambda \mu}g^{\mu\nu}(\partial_\nu)_\flat = \delta^\nu_{\lambda}(\partial_\nu)_\flat = (\partial_\lambda)_\flat \implies \partial_\lambda = (g_{\lambda\mu}{\rm d}x^\mu)^\sharp = g_{\lambda\mu}({\rm d}x^\mu)^\sharp.$$(Relabel $\mu \to \nu$ and $\lambda \to \mu$ in this order if it makes you more comfortable in the last equality above.)
A: 
SUMMARY. This is formally wrong, as Ivo already showed. If the metric is $\delta_{\mu \nu}dx^\mu\otimes dx^\nu$, then it is morally correct. Otherwise, it is not; see the example at the end of this post.

My understanding of these things is that you can raise or lower indices in the coordinates of a tensor, not in the tensor itself. To explain what I mean, let $\mu_0\in \{1, 2, \ldots n\}$ be fixed. The 1-form $dx^{\mu_0}$ is the tensor whose coordinates in the basis $dx^1, dx^2, \ldots, dx^n$ are 
$$
(\delta^{\mu_0}{}_{\nu}\ :\ \nu=1, 2, \ldots, n), $$
and we denote them by $\delta^{\mu_0}{}_{\nu}$. (This $\delta$ is the Kronecker symbol, which equals $1$ if both indices agree and $0$ otherwise. The spacing in the indices is not important). 
Dually, the vector field $\frac{\partial}{\partial x^{\mu_0}}$ is the tensor whose coordinates are $\delta_{\mu_0}{}^\nu$. And this is the end of the story unless we introduce a Riemannian metric. 
If we introduce a such metric $g_{\mu\nu}$ (or $g_{\mu\nu}dx^\mu\otimes dx^\nu$, if you prefer the extended version), then we can raise or lower indices by contracting with this metric, which is another definition of the "musical isomorphisms" of Ivo's answer. In the case of the tensor we introduced previously, if $g_{\mu\nu}=\delta_{\mu\nu}$, then raising the second index in the first tensor we obtain the second; 
$$
\delta^{\mu_0}{}_\nu \delta^{\nu \rho}=\delta^{\mu_0\rho}. $$ 

WARNING! If the metric is not $\delta_{\mu \nu}dx^\mu\otimes dx^\nu$, then the above fails, and (in the language of Ivo's answer) it is not true that 
$$
\left(\frac{\partial}{\partial x^\mu}\right)_b =dx^\mu, \quad \frac{\partial}{\partial x^\mu}=(dx^\mu)^{\#}.$$ 
Let me make an explicit example: parametrize $\mathbb S^2$ (minus a semicircle) as 
$$
(\cos \theta, \sin \theta\cos \phi), \qquad \text{where }\theta\in(0, \pi),\phi\in(0, 2\pi).$$
Then the metric tensor of $\mathbb S^2$ is 
$$
g = d\theta^2 + \sin^2\phi d\phi^2,$$
hence the matrix $g_{\mu \nu}$ is 
$$
\begin{bmatrix} 1 & 0 \\ 0 & \sin^2\phi\end{bmatrix}. $$
In particular, $g_{\mu \nu}$ is not the Kronecker symbol. 
Now, let us compute $(d\phi)_b$. The components of this tensor, in the basis $d\theta, d\phi$, are $(0, 1)$. We should contract then with the metric, which amounts to compute 
$$
(0, 1)\begin{bmatrix} 1 & 0 \\ 0 & \sin^2\phi\end{bmatrix} = (0, \sin^2\phi).$$ 
Hence 
$$
(d\phi)_b= \sin^2\phi \frac{\partial}{\partial \phi}.$$
