Nijenhuis tensor of a complex manifold Using the fact that for any smooth vector fields $X,Y:TM\rightarrow M$ where $(M,J)$ is an almost complex manifold, then $N(fX,Y)=fN(X,Y)$ for any smooth function $f$ on $M$. Where $N(X,Y)=J[X,Y]-[JX,Y]-[X,JY]-J[JX,JY]$  is the Nijenhuis tensor.
How can we prove that if $M$ is a complex manifold then $N\equiv0$?
 A: If $M$ is a complex manifold, let $(z_1,...,z_n)$ be a local complex coordinates at the point $p$ of  $M$. If $z_i=x_i+\sqrt{-1}y_i$ for $i=1,2,...,n$, then as a smooth manifold, $M$ has local coordinates $(x_1,...,x_n,y_1,...,y_n)$ and the tangent space at $p$ is spanned by 
$\displaystyle\frac{\partial}{\partial x_i}$ and $\displaystyle\frac{\partial}{\partial y_i}$ for $i=1,2,...,n.$
Define $J:T_pM\to T_pM$ by 
$$J(\frac{\partial}{\partial x_i})=\frac{\partial}{\partial y_i}\mbox{ and }
J(\frac{\partial}{\partial y_i})=-\frac{\partial}{\partial x_i}\mbox{ for all }i=1,2,...,n.$$
Then $J^2=-id$ which shows that $J$ is an almost complex structure. 
With this almost complex structure $J$, we can show that 
$$N(\frac{\partial}{\partial x_i}, \frac{\partial}{\partial x_j})
=N(\frac{\partial}{\partial y_i}, \frac{\partial}{\partial y_j})
=N(\frac{\partial}{\partial x_i}, \frac{\partial}{\partial y_j})=
N(\frac{\partial}{\partial y_i}, \frac{\partial}{\partial x_j})=0,
$$
since we always have 
$$[\frac{\partial}{\partial x_i}, \frac{\partial}{\partial x_j}]=
[\frac{\partial}{\partial y_i}, \frac{\partial}{\partial y_j}]=
[\frac{\partial}{\partial x_i}, \frac{\partial}{\partial y_j}]=0.$$
From this, we have $N(X,Y)=0$ since $N$ is linear and $N(fX,Y)=fN(X,Y)$ as you have said. 
