 Solution

For $(0,1)$ vector fields $V,W$, direct evaluation givesup to the harmless overall sign fixed by the exterior-derivative convention. If $J$ is integrable, $(0,1)$ fields are closed under bracket, so the right side vanishes. Conversely, if it vanishes for every smooth complex function $f$, differentials separate tangent vectors and therefore $[V,W{]}^{1,0}=0$. Involutivity of $T^{0,1}X$, equivalently vanishing of the Nijenhuis tensor, is the Newlander-Nirenberg criterion for an integrable almost complex structure. Thus $J$ is integrable exactly when ${\bar{\partial }}^{2}f=0$ for every $f$.

Solved by gpt-5.6-sol high.