How to interpret $\nabla_x (f(u(x)))$? Is $\nabla_x (f(u(x)))= (\nabla f) (u(x)) \nabla u$ true? 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, composition of $f$ and $u$ is: $f(u)= |u|^{\frac{4}{n-2}}u.$ 


  My Question is: How should I interpret  $\nabla_x (f(u(x)))$? Is  $\nabla_x (f(u(x)))= (\nabla f) (u(x)) \nabla u$ true? 


[Is this a standard notaions? My understnding for the gradient is: if $g:\mathbb R^n\to \mathbb C$ nice function then $\nabla g (x) = (\frac{\partial g(x)}{\partial x_1},..., \frac{\partial g(x)}{\partial x_n}), x=(x_1,...,x_n)\in \mathbb R^n$. Or am I missing something here?  But I do not knwo what is means $\nabla_x g$, which was used by the autors in the remark mentioned below]

Motivation: I have been trying to figure out  Remark 2.3 here See also this question
When you are starting with this work, it is best to look at the components.  Let $x= (x_1,x_2,...,x_n)$. $\nabla_x(f(u(x)))$ is an $n-$component vector.  The first component of $\nabla_x(f(u(x)))$ is $$\frac{\partial f}{\partial x_1}=\frac{d f}{du}\frac{\partial u}{\partial x_1}$$
The first derivative in this chain rule expression is not a partial derivative, since $f$ is a function of a scalar $u$ (so you won't have anything that looks like $\nabla f$).  Since the other components will be similar, you obtain
$$\nabla_x(f(u(x)))=(\frac{d f}{du}\frac{\partial u}{\partial x_1},\frac{d f}{du}\frac{\partial u}{\partial x_2},...,\frac{d f}{du}\frac{\partial u}{\partial x_n})$$ 
$$=\frac{df}{du} \nabla u$$
Try the chain rule for multivariate calculus: https://www.math.hmc.edu/calculus/tutorials/multichainrule/