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 $N\equiv0$?
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.