 Solution

Conversely, suppose the exterior derivative of every $(1,0)$-form has no $(0,2)$ part. For sections $U,V$ of $T^{0,1}M$, the same exterior derivative formula givesThe common annihilator of all $(1,0)$-covectors is $T^{0,1}M$. Local $(1,0)$-coframes therefore show that $[U,V]$ belongs to $T^{0,1}M$. Conjugating gives bracket closure of $T^{1,0}M$. Thus the two criteria are equivalent. These are the bracket and differential-form criteria for integrability, and establish integrability in the sense used here.