I am not an expert but the question:
"Does there exist a simply-connected closed Riemannian Ricci flat $n$-manifold with $SO(n)$-holonomy?"
is a well-known open problem. Note that Schwarzschild metric is a complete Ricci flat metric on $S^2\times\mathbb R^2$ with holonomy $SO(4)$, so the issue is to produce compact examples; I personally think there should be many. The difficulty is that it is hard to solve Einstein equation on compact manifolds. If memory serves me, Berger's book "Panorama of Riemannian geometry" discusses this matter extensively.
The whole point is that if the metric is Kähler, then the Chern connection coincides with the complexification of Levi-Civita connection of the riemanian metric on the underlying real manifold given by the real part of the Kähler hermitian metric.
Moreover, this happens if and only if the starting metric is Kähler (and is indeed my favorite definition of Kähler metric).
So, let me give you a hint.
If $\omega$ is any hermitian metric on $X$, then $\det\omega$ is the induced metric on $\det T_X$ (this line bundle is usually referred as to the anticanonical bundle $K_X^{*}$). The Chern curvature $\Theta$ of this metric is given by
$$
i\Theta(K_X^*,\det\omega)=-i\partial\bar\partial\log\det\omega.
$$
Now, in general, the Chern curvature (with respect to the induced hermitian metric) of the determinant $\det E$ of a hermitian holomorphic vector bundle $(E,h)$ is given by the trace of the Chern curvature of $E$:
$$
i\Theta(\det E,\det h)=i\operatorname{Tr}\Theta(E,h).
$$
Therefore,
$$
-i\partial\bar\partial\log\det\omega=i\operatorname{Tr}\Theta(T_X,\omega).
$$
But $\Theta(T_X,\omega)$ is the complexification of the curvature $R^{LC}$ of the Levi-Civita connection of $X$ $\nabla^{LC}_{X,\Re\omega}$ with respect to $\Re\omega$, where $\Re\omega$ is the induced riemannian metric on the underlying real manifold.
Finally, by definition, the Ricci curvature of $(X,\Re\omega)$ is the trace of $R^{LC}$.
There is a small missing point in my sketchy argument: the riemannian Ricci curvature is a symmetric $2$-controvariant tensor, while the Ricci form is a closed real $(1,1)$-form. You have to show that in fact the riemannian Ricci curvature of (the real part of) a Kähler metric is invariant with respect to the complex structure $J$, then you apply the $1-1$ correspondence between real $(1,1)$-forms and symmetric $J$-invariant $2$-controvariant tensors.
Best Answer
The point is that a Bochner formula shows that, if the Kähler metric is Ricci-flat and the manifold is compact, then every global holomorphic $p$-form must be parallel.
In particular, if you have a complex, compact symplectic manifold that is Kähler, then you have a nonvanishing holomorphic volume form (the top power of the holomorphic symplectic $2$-form). You can then use Yau's theorem to find a Kähler metric whose volume form is a constant multiple of this holomorphic volume form wedged with its conjugate and this metric will have vanishing Ricci curvature. Then, by the Bochner formula, the symplectic form has to be parallel with respect to the Levi-Civita connection of the Kähler metric. In particular, this implies that the holonomy preserves this form and hence is a subgroup of $\mathrm{Sp}(n,\mathbb{C})\cap\mathrm{U}(2n)=\mathrm{Sp}(n)$. If you know, for some reason, that the manifold is simply-connected and not a product, then deRham's Theorem shows that the holonomy must act irreducibly. Berger's classification of the possible irreducibly acting holonomy groups then implies that the metric must have restricted holonomy equal to $\mathrm{Sp}(n)$, since no proper subgroup of this group can act irreducibly and be the holonomy.