Let $f(u)= |u|^{\frac{4}{n-2}}u.$ Then $ |\partial_z f(u) | \leq C |u|^{\frac{4}{n-2}}$? Let $u:\mathbb R^{n+1} \to \mathbb C$ be a function and $f:\mathbb C \to \mathbb C$ such that $f(z)=|z|^{\frac{4}{n-2}}z.$ Thus $f(u)= |u|^{\frac{4}{n-2}}u.$

My Question is: Can we say $ |\partial_z f(u) | \leq C |u|^{\frac{4}{n-2}}$ and $ |\partial_{\bar{z} }f(u) | \leq C |u|^{\frac{4}{n-2}}$ ($C$ is some constant)? If yes,  how to show it.
Can we say that : $ | \partial_z f(u)-  \partial_z f(v) | \leq C |u-v| (|u|^{\frac{6-n}{n-2}} + |u|^{\frac{6-n}{n-2}})$ for  $3 \leq n \leq 6$?

Edit: I think $f(u)$ is the composition of functions $f$ and $u$. But it is not clear to me that how I should interpret $\partial_z f(u)$.
Motivation: This is nonlinearity occur in nonlinear Schrodinger equation, see for Remark 2.3
Edit 2: From the argument of  H. H. Rugh, we have
$$ ||u|^p - |v|^p| \leq C_p \ |u-v|\ \left| |u|^{p-1} + |v|^{p-1}\right|$$
And we may take $p=4/(N-2)$, but then how to get
$|\partial_{\bar{z}}f(u)- \partial_{\bar{z}}f(v)| \leq | |u|^{4/(N-2)}-|v|^{4/(N-2)}|$?
In other words how to conclude the second inequality?
Please would you also let know me know how this follows:
How to show:

(1)$(\nabla f) (u(x))=  \partial_{z} f(u(x)) u(x)  + \partial _{\bar{z}}f(u(x)) \overline{u(x)}$ (A)
(2) How to use (A) to get  $|f(u)-f(v)| \leq |u-v| (|u|^{4/(N-2)} + |v|^{4/(N-2)})$
(3) How should I interpret  $\nabla_x (f(u(x)))$? Is  $\nabla_x (f(u(x)))= (\nabla f) (u(x)) \nabla u$ true? (4)  What is relation between $\partial_z f(u)$ and $\nabla f) (u(x))$ ? Is it same?

 A: Ok, I got the point from the article of Kenig and Merle. They seem to consider $f(z)=|z|^{\frac{4}{N-2}} z = (z\; \bar{z})^{\frac{2}{N-2}}z=z^{\frac{N}{N-2}} \; \bar{z}^{\frac{2}{N-2}}$ as a function of $z$ and $\bar{z}$. Then e.g.
$$ \partial_z f(z)= \frac{N}{N-2}  \; z^{\frac{2}{N-2}} \; \bar{z}^{\frac{2}{N-2}} , \ \ \ |\partial_z f(z)| \leq \frac{N}{N-2} |z|^{\frac{4}{N-2}}$$
(and then insert $u$ to get the stated bound).
For the second they use that for real positive numbers $a,b$ and $p\geq 1$:
$$ |a^p - b^p| \leq C_p \ |a-b|\ \left|a^{p-1} + b^{p-1}\right|  \ \ \ (*)$$
Then insert $a=|u|$, $b=|v|$ and use that
Using the above as well as $ \left||u|-|v|\right|\leq |u-v|$ you may then estimate: 
$$ \left| |u|^p u - |v|^p v\right| \leq |u|^p |u-v| + |v| \left| |u|^p-|v|^p \right| \ \ \ (**)$$
In their case $N\leq 6$ is because we want $p\geq 1$ and the estimate in any case  only makes sense when $N\geq 3$.
I don't understand (1) [to me $\nabla f$ is a vector?] but you don't need it to show (2) except that I think it is wrong as stated. It lacks a prefactor in general greater than 1 (but this suffices for the rest). You may use the estimate  $(*)$ and $(**)$.
We also have:
$$ \partial_\bar{z} f(z)= \frac{2}{N-2}  \; z^{\frac{N}{N-2}} \; \bar{z}^{\frac{4-N}{N-2}} , \ \ \ |\partial_\bar{z} f(z)| \leq \frac{2}{N-2} |z|^{\frac{4}{N-2}}$$
