For ease of reading, I'm listing all my definitions at the bottom of this question.
We let $X$ be an $n$-dimensional complex manifold. Then we can define the sheaf of holomorphic functions $\mathcal{O}_X$ on $X$. Fix $x\in X$, then we have the local ring $\mathcal{O}_{X,x}$, with maximal ideal $\mathfrak{m}_{X,x}$. Note that if we define the (surjective) evaluation map $$\theta_x:\mathcal{O}_{X,x}\to\mathbb{C},\ \langle U,f\rangle\mapsto f(x)$$ then $\ker\theta_x=\mathfrak{m}_{X,x}$, and $\theta_x$ is surjective, so we can identify $\mathcal{O}_{X,x}\big/\mathfrak{m}_{X,x}$ with $\mathbb{C}$.
A $\mathbb{C}$-linear map $v:\mathcal{O}_{X,x}\to\mathbb{C}$ is said to be a derivation at $x$ if it satisfies the product rule $$v(fg)=\theta_x(f) v(g) + \theta_x(g) v(f)$$ and the set $\mathrm{T}_x X$ of derivations at $x$ has a natural vector space structure.
Claim: $$(T_x X)^*\cong \mathfrak{m}_{X,x}\big/\mathfrak{m}_{X,x}^2$$
I think I have reduced to the case where $X=\mathbb{C}^n$ and $x=0$. However, I've hit a snag in trying to calculate $\mathfrak{m}_{X,x}\big/\mathfrak{m}_{X,x}^2$. My first observation was that we can naturally embed $$\mathcal{O}_{\mathbb{C}^n,0}\hookrightarrow\mathbb{C}[[z_1,\ldots,z_n]]$$ and that, using multi-index notation, we have that $$\mathfrak{m}_{X,x} =\left\{\sum_{|\nu|\ge 1} a_\nu \mathfrak{z}^\nu\in\mathbb{C}[[z_1,\ldots,z_n]]\ \bigg\vert\ \sum_{|\nu|\ge 1} a_\nu \mathfrak{z}^\nu\in\mathcal{O}_{X,x}\right\}$$ Now, if I can prove that $$\mathfrak{m}_{X,x}^2 =\left\{\sum_{|\nu|\ge 2} a_\nu \mathfrak{z}^\nu\in\mathbb{C}[[z_1,\ldots,z_n]]\ \bigg\vert\ \sum_{|\nu|\ge 2} a_\nu \mathfrak{z}^\nu\in\mathcal{O}_{X,x}\right\}$$ then we can identify $\mathfrak{m}_{X,x}\big/\mathfrak{m}_{X,x}^2$ with $$\left\{\sum_{|\nu|=1} a_\nu \mathfrak{z}^\nu\in\mathbb{C}[[z_1,\ldots,z_n]]\ \bigg\vert\ \sum_{|\nu|=1} a_\nu \mathfrak{z}^\nu\in\mathcal{O}_{X,x}\right\}\cong\mathbb{C}^n\cong\mathrm{T}_0 \mathbb{C}^n.$$
So, I have reduced my problem essentially to showing that for $$f\in\left\{\sum_{|\nu|\ge 2} a_\nu \mathfrak{z}^\nu\in\mathbb{C}[[z_1,\ldots,z_n]]\ \bigg\vert\ \sum_{|\nu|\ge 2} a_\nu \mathfrak{z}^\nu\in\mathcal{O}_{X,x}\right\}$$ defined on some neighborhood of the origin, we can define $\sqrt{f}$ as a holomorphic function in some neighborhood of the origin. This seems like it must be some simple exercise in the theory of complex variables, but I guess I just don't know the necessary theorems to prove this.
Alternatively, is there a more convenient way of proving that these two definitions are essentially equivalent?
Definition: A topological space $X$ is an $n$-dimensional complex manifold if it is endowed with a set $\mathscr{A}$ (called a holomorphic structure) such that
$\mathscr{A}$ has elements of the form $(U,\varphi)$, where $U\subseteq X$ is an open subset, $\varphi:U\to\varphi(U)$ is a homeomorphism, and $\varphi(U)\subseteq\mathbb{C}^n$ is open
for any two overlapping charts $(U,\varphi)$ and $(V,\psi)$, $$\psi\circ\varphi^{-1}\vert_{\varphi(U\cap V)}:\varphi(U\cap V)\to\psi(U\cap V)$$ is holomorphic with holomorphic inverse.
Definition: For any open subset $U\subseteq X$, a map $f:U\to\mathbb{C}$ is said to be holomorphic if for every holomorphic chart $(V,\varphi)\in\mathscr{A}$ with $V\subseteq U$, $$f\circ\varphi^{-1}:\varphi(V)\to\mathbb{C}$$ is holomorphic. The set of holomorphic functions on $U$ is denoted by $\mathcal{O}_X(U)$, and is in fact a ring. This makes $\mathcal{O}_X$ into a sheaf, so we can define the stalk $\mathcal{O}_{X,x}$ of $\mathcal{O}_X$ at $x\in X$, which is a local ring. Therefore, $(X,\mathcal{O}_X)$ is a locally-ringed space.