How do you differentiate $f(x_1,x_2,...,x_n)=(f_1(x_1,x_2,...,x_n), f_2(x_1,x_2,...,x_n), ..., f_n(x_1,x_2,...,x_n))$? How do you differentiate $$f(x_1,x_2,...,x_n)=(f_1(x_1,x_2,...,x_n), f_2(x_1,x_2,...,x_n), ..., f_n(x_1,x_2,...,x_n))$$?
Is it by taking the derivative of each of the component functions and then
$$f'(x_1,x_2,...,x_n)=(f_1'(x_1,x_2,...,x_n), f_2'(x_1,x_2,...,x_n), ..., f_n'(x_1,x_2,...,x_n))$$
?
How do you apply the chain rule to such vector?
 A: You have a function $f:\mathbb{R}^n\to\mathbb{R}^n$ and functions $f_i:\mathbb{R}^n\to\mathbb{R}$.
Generally, the derivative of a function $f:\mathbb{R}^n\to\mathbb{R}^m$ at a point $a$ is, when it exists, a linear application $L:\mathbb{R}^n\to\mathbb{R}^m$ such that $\frac{\|r(h)\|}{\|h\|}\to0\quad(h\to0)$, where $f(a+h)=f(a)+L(h)+r(h)$. Then, there is a matrix $A\in M_{m\times n}$ such that $L(h)=Ah$ and we write $f'(a):=L$ (seen as a function) or $f'(a):=A$ (seen as a matrix associated with $L$).
Hence, the derivative of $f$ is an $n\times n$ matrix and the derivatives of the $f_i$ are $1\times n$ matrices.
A theorem says that if the derivative of $f$ exists at $a$, then the jacobian matrix $J_f(a)$ of $f$ at $a$ exists and $f'(a)=J_f(a)$ (the converse is false).
All in all, what you wrote is incorrect but would be correct if you put it like this:
$$
f'(x_1,x_2,\ldots,x_n)=\begin{pmatrix}f_1'(x_1,x_2,\ldots,x_n)\\f_2'(x_1,x_2,\ldots,x_n)\\\cdots\\f_n'(x_1,x_2,\ldots,x_n)\end{pmatrix},
$$
each $f'_i$ representing rows in the matrix.
A: I presume the functions $f_i,f$ all map from $\mathbb{R}^n$ into $\mathbb{R}$. In this case you can either consider the partial derivatives with respect to a given variable $\partial_if$, or compute the total derivative $Df$. In the first case the formula is given by the chain rule:
$$\partial_i f=\sum_{j=1}^n\partial_{f_j}f\cdot \partial_i f_j.$$
In the second $Df$ is just an $1\times n$ vector containing all partial derivatives calculated just as previously shown, i.e
$$Df=(\partial_1f,\ldots,\partial_nf)$$
