I am a physics student and I am currently taking a course on GR, but the module is not presented in a particularly rigorous manner. My problem lies in one line of the online notes (which unfortunately I cannot share), that reads: $$0 = \partial_{\mu}(\textbf{e}_{\nu} \textbf{e}^{\nu}) = \textbf{e}_{\nu} \partial_{\mu} \textbf{e}^{\nu} + \textbf{e}^{\nu} \partial_{\mu} \textbf{e}_{\nu} \tag{A}$$
I hope that the reader familiar with GR/Differential Geometry finds this trivial (again, I cannot share the notes).
So, the equation is $\textbf{NOT}$ written in an environment where all objects have been carefully defined (the notes are mostly intuition based) so I will try to build things from scratch so that the readers can see my frustrations.
Consider a smooth manifold $\mathcal{M}$ of dimension 4 (here, we define naturally the index set $\mathbb{I}_4 := \{1,2,3,4\}$) with an atlas $(U, \varphi)$. At each point $x\in \mathcal{M},$ there exists a vector space of all derivations at $x$ we call the tangent space $T_x \mathcal{M}$. The space is spanned by the basis
$$\mathbb{B}_x := \{\textbf{e}_{\mu}(x); \mu \in \mathbb{I}_4 \}$$
We can then also consider the cotangent vector space at $x \in \mathcal{M}$, spanned by the basis $$\bar{\mathbb{B}}_x := \{\textbf{e}^{\mu}(x); \mu \in \mathbb{I}_4 \}$$ The linear functionals have been defined "naturally" with the biorthogonality property such that: $$\textbf{e}^{\mu}(x) \left[ \textbf{e}_{\kappa}(x)\right] = \delta^{\mu}_{\ \kappa} \text{ for all } \mu, \kappa \in \mathbb{I}_4 \text{ and } x \in \mathcal{M}$$ With a trivial abbreviation, the last equation reduces to the classic: $$\textbf{e}^{\mu} \textbf{e}_{\kappa} = \delta^{\mu}_{\ \kappa}$$ Where we arrive at my first problem, according to these (as far as I am concerned, standard definitions) the notation $\textbf{e}_{\nu} \textbf{e}^{\nu}$ in (A) is not even defined. Could this be a case of just bad notation or is there something else (apart from the trivial abbreviation $\textbf{e}_{\nu} \textbf{e}^{\nu} := \textbf{e}^{\nu} (\textbf{e}_{\nu})$). Second and most importantly, how is the object $\partial_{\mu}$ defined? If I accept that $\textbf{e}_{\nu} \textbf{e}^{\nu} := \textbf{e}^{\nu} (\textbf{e}_{\nu})$ and equation (A), then $\partial_{\mu}$ seems to be an object that acts on scalars (members of $\mathbb{R}$), vectors ($\in T_{x}\mathcal{M}$) and co-vectors ($\in T_{x}\mathcal{M}^{*}$). There seems to be a lot of more primitive, and hidden, definitions lying here and I would very much appreciate some insight.
A few things:
First, notationally, I am assuming $e_\nu$ is a vector field and $e^\nu$ is a 1-form. There are two conventions physicists use, and I'm using the one consistent with normal differential geometry. Second, I'm assuming you're working with a Riemannian manifold.
From context, it is clear that $e_\nu$ is a frame. It is not clear if $e_\nu$ is an orthonormal frame, or one derived from a choice of local coordinates. In flat space, we commonly conflate the two but it's important to separate out these two concepts for curved spaces.
Either way, I would interpret your $e_\nu e^\nu $ as a trace over the $(1,1)$ tensor $e_\nu\otimes e^\rho$. In either of the above interpretations, you get a constant function equal to the dimension of your manifold, so taking partial derivative $\partial_\mu$ will get you $0$.
The real question is how to interpret $\partial_\mu e_\nu$ and $\partial_\mu e^\nu$ since the first one is taking a derivative of a vector field and the second is taking a derivative of a 1-form. On a Riemanninan manifold, your metric induces a unique connection $\nabla$ called the Levi-Civita connection. You should think of this as a generalization of a partial derivative, but you may take the partial derivative of any tensor, and it returns a tensor of the same type. Therefore, it looks to me like you should interpret $\partial_\mu e_\nu$ as the vector field $\nabla_{\partial_\mu}e_\nu$, and $\partial_\mu e^\nu$ as the 1-form $\nabla_{\partial_\mu}e^\nu$. Furthermore, $\nabla$ satisfies a product rule wrt tensor product, namely $$\nabla_{\partial_\mu}(e_\nu\otimes e^\rho)=(\nabla_{\partial_\mu}e_\nu)\otimes e^\rho +e_\nu \otimes (\nabla_{\partial_\mu}e^\rho).$$ One can show taking trace commutes with $\nabla$, so that \begin{align*} 0&=\nabla_{\partial_\mu}\text{Tr}(e_\nu\otimes e^\rho)\\ &=\text{Tr}[\nabla_{\partial_\mu}(e_\nu\otimes e^\rho)]\\ &=\text{Tr}[(\nabla_{\partial_\mu}e_\nu)\otimes e^\rho] +\text{Tr}[e_\nu \otimes (\nabla_{\partial_\mu}e^\rho)]\\ \end{align*} where in the first equality I use that $\nabla_{\partial_\mu}f=\partial_\mu f$ for any scalar field $f\in C^\infty(M)$. The last equality, I'm guessing, is what is meant by $(\partial_\mu e_\nu) e^\nu+e_\nu(\partial_\mu e^\nu)$.