It is easier to prove this result for 1-forms, instead of vector fields. On (1,0)-forms, $\nabla^{0,1}=\bar\partial$ because the Levi-Civita connection is torsion-free, hence $\bigwedge(\nabla(\eta))=d\eta$, where $\bigwedge:\; \Lambda^1M \otimes \Lambda^k M \to \Lambda^{k+1}(M)$ is the exterior product. This is actually true for all torsion-free connection, not necessarily the Levi-Civita. From   $\nabla^{0,1}=\bar\partial$ we obtain that $(0,1)$-part of Levi-Civita connection applied to (1,0)-forms is equal to $(0,1)$-part of Chern connection applied to (1,0)-forms. Now, the Chern connection is by definition the only connection on $\Lambda^{1,0}(M)$ which has this property and preserves the Hermitian metric. Since the Levi-Civita connection preserves the Hermitian metric, it is equal to Chern connection.