How to prove a function which is polynomial in the coordinates is differentiable everywhere The question is:
Using the chain rule to prove that a function $f:\mathbb{R}^n\rightarrow \mathbb{R}$ which is polynomial in the coordinates is differentiable everywhere.
(The chain rule is for the use under function composition circumstances, how to apply it here to prove that the function f which is polynomial in the coordinates is differentiable everywhere?)
 A: The question is interesting. First of all, the question @Kaster poses is relevant: what do you mean by polynomial in the coordinates? For now, we'll assume the same as he did, namely that each monomial occuring in $f$ contains at most one variable $x_i$.
From now on, we will assume $f$ has degree $d$ and we will write 
$$f(x_1,\ldots,x_n) = \sum_{i=1}^{n}\sum_{j=0}^{d} a_{ij} x_i^j .$$
The question asks you to use the chain rule in order to prove it's differentiable. This seems a bit elaborate, but basically asks to split this map in parts (i.e. $f=f_r\circ f_{r=1}\circ\ldots\circ f_1$) of which we already know they are differentiable. It then follows from the chain rule that their composition $f$ is differentiable.
As mentioned, this may be somewhat elaborate. We will just assume that $x\mapsto x^m$ and $x\mapsto ax$ are known to be differentiable ($m\in\mathbb{N}$, $a\in\mathbb{R}$) and that matrix multiplication is differentiable as well (if you don't like this, you could use the sum rule instead). The map $f$ can then be split as
$$(x_1,\ldots,x_n)\mapsto \left(\begin{array}{cccc}1 & x_1 & \ldots & x_1^d\\\vdots & \ddots & & \vdots\\\vdots& & \ddots & \vdots\\1 & x_n & \ldots & x_n^d\end{array}\right) \mapsto \left(\begin{array}{cccc}1\cdot a_{10} & a_{11}x_1 & \ldots & a_{1d}x_1^d\\\vdots & \ddots & & \vdots\\\vdots& & \ddots & \vdots\\1\cdot a_{n0} & a_{n1}x_n & \ldots & a_{nd}x_n^d\end{array}\right)\\
\mapsto(1,\ldots,1)\cdot \left(\begin{array}{cccc}1\cdot a_{10} & a_{11}x_1 & \ldots & a_{1d}x_1^d\\\vdots & \ddots & & \vdots\\\vdots& & \ddots & \vdots\\1\cdot a_{n0} & a_{n1}x_n & \ldots & a_{nd}x_n^d\end{array}\right) \cdot \left(\begin{array}{c}1\\\vdots\\\vdots\\1\end{array}\right). $$
If you ensure the $a_{ij}$ are numbered correctly (if your convention seems to be wrong, you may try to transpose everything) and the correct number of ones in the final step, the composite is well-defined and corresponds to $f$. Moreover, it only contains steps of the form we assumed to know to be differentiable and therefore $f$, their composite, is differentiable as well.
I hope this is more or less what you wanted/needed. A few remarks need to be made, though.
Remark 1. All the coefficients with $j=0$ correspond to the constant term of $f$. You may take all of them, possibly except for one, to be zero if you like, but this is not necessary.
Remark 2. I started by assuming a meaning of "polynomial in coordinates". There are other things you could've meant. The first two things I tried, however, contradicted your statement. If you meant $f(0,\ldots,0,x_i,0,\ldots,0)$ to be a polynomial for each $i$, a counterexample is given by $n=2$, $f(x,y)= |x|y$. If you meant that $f(1,\ldots,1,x_i,1,\ldots,1)$ is polynomial for all $i$, $n=2$, $f(x,y)=(1-y)|x|$ is a counterexample.
