Basically the question is here Analytification of algebraic differential forms for smooth complex projective scheme of finite type $X$. Let $h: X^{an} \rightarrow X$ be the analytification morphism. I would love to see a more detailed proof why the pullback of $\Omega_{X/\mathbb{C}}$, sheaf of Kähler differentials of $X$, is $\Omega^{an}$, the sheaf of holomorphic 1-forms of $X^{an}$. I think we just need to construct a canonical morphism $h^*\Omega_{X/\mathbb{C}} \rightarrow \Omega^{an}$ and prove that it is injective or surjective, then the isomorphism follows from the fact that both are locally free of the same rank.
Thank you in advance.
Consider smooth scheme $X$ over $\mathbb{C}$. Fix a closed point $x\in X$ and an open affine neighborhood $U$ of $x$ in $X$. Let $\mathbb{C}[U]$ be the coordinate ring of $U$. An ideal $\mathfrak{m}_x$ of $x$ in $C[U]$ is maximal and it admits a set of generators $z_1,...,z_n$ such that their classes in $\mathfrak{m}_x/\mathfrak{m}^2_x$ are linearly independent. We define a morphism of $\mathbb{C}$-algebras
$$\mathbb{C}[t_1,...,t_n]\rightarrow C[U]$$
such that $t_i \mapsto z_i$ for every $i=1,2,...,n$. This morphism corresponds to a morphism of $\mathbb{C}$-schemes
$$\phi:U\rightarrow \mathbb{A}^n_{\mathbb{C}}$$
given by functions $z_1,...,z_n$. Since $z_1,...,z_n$ form a minimal set of generators of the maximal ideal $\mathfrak{m}_x$ and $U$ is smooth, we derive that $\phi$ is an étale morphism on some open affine neighborhood $V$ of $x$ in $U$ (this is just an invocation of the fact that étale locus is Zariski open). Hence we obtain an étale morphism $\phi_{\mid V}:V\rightarrow \mathbb{A}^n_{\mathbb{C}}$.
Let $i:V^{\mathrm{an}}\rightarrow V$ be the canonical morphism of $\mathbb{C}$-ringed spaces.
Now we make two observations
The sheaf $\Omega_V$ of algebraic differential $1$-forms is a free sheaf generated by $dz_1$,...,$dz_n$ (this follows from the fact that $\phi_{\mid V}$ is étale).
The analytification ${\phi_{\mid V}}^{\mathrm{an}}:V^{\mathrm{an}}\rightarrow \mathbb{C}^n$ is locally biholomorphic (because it is analytification of an étale morphism) and given by $i^{\#}(z_1),...,i^{\#}(z_n)$ ($z_1,...,z_n$ viewed as holomorphic functions on $V^{\mathrm{an}}$). Hence $di^{\#}(z_1),...,di^{\#}(z_n)$ generate freely the sheaf $\Omega_{V^{\mathrm{an}}}$ of holomorphic differential forms on $V^{\mathrm{an}}$
Observations 1) and 2) imply that the analytification $$(\Omega_V)^{\mathrm{an}} =i^*\Omega_V = \left(i^{-1}\Omega_V\right)\otimes_{i^{-1}\mathcal{O}_V} \mathcal{O}_{V^\mathrm{an}}$$ of algebraic $1$-forms is a free $\mathcal{O}_{V^\mathrm{an}}$-sheaf generated by $i^*(dz_1),...,i^*(dz_n)$ and the sheaf of holomorphic $1$-forms $\Omega_{V^{\mathrm{an}}}$ is free over $\mathcal{O}_{V^\mathrm{an}}$ and generated by $di^{\#}(z_1),...,di^{\#}(z_n)$. Since the canonical morphism $$i^*\Omega_V\rightarrow \Omega_{V^{\mathrm{an}}}$$ is given by $i^*(df)\mapsto di^{\#}(f)$, we derive that $i^*\Omega_V\rightarrow \Omega_{V^{\mathrm{an}}}$ is an isomorphism of sheaves of $\mathcal{O}_{V^{\mathrm{an}}}$-modules. Since this morphism of sheaves is the restriction to $V^{\mathrm{an}}$ of a morphism $$j^*\Omega_X\rightarrow \Omega_{X^{\mathrm{an}}}$$ induced by the analytification morphism $j:X^{\mathrm{an}}\rightarrow X$, we derive that $j^*\Omega_X\rightarrow \Omega_{X^{\mathrm{an}}}$ is an isomorphism.