Absolute values equivalence in $\mathbb{F}_{q}(x)$ Let $q=p^n$ a prime power, $f(x)$ a monic irreducible polynomial in $\mathbb{F_q}[x]$ and $q_f = \text{the size of the residual field of f.}$
Define the following absolute values over $\mathbb{F}_q(x)$:
$|g(x)|_{f(x)} := q_f^{v_f}$ where $v_f$ is the greatest power of $f$ that divides g.
$|g(x)|_\infty := q^{\text{deg}(g(x))}$.
I am asked to show that if I am given an absolute value such that $|x| > 1$ then $| \cdot|$ is equivalent to $|\cdot|_\infty$ and if $|x| \leq 1$ then $|\cdot|$ is equivalent to $|\cdot|_{f(x)}$ for some $f$. 
I am not really clear on what the definition of the residual field of $f$ is, but I would assume that it is $\mathbb{F}_q[x] / \langle f(x)\rangle$. 
Also, I know that if $|x| = t > 1$ then $|x| = |x|_\infty ^c=q^c$ where $c=\frac{ln(t)}{ln(q)}$ but I don't know how to generalize this for any polynomial.  
 A: Let $|\cdot|$ be an absolute value on $\mathbb{F}_q(x)$. I will show that it is equivalent to one of the options that you give, or to the trivial absolute value (the trivial value is: $|g| = 1$ for all non-zero $g$ and $|0| = 0$). Actually, I will only show this on the polynomial ring $\mathbb{F}_q[x]$, but that is, strictly speaking, where your definitions make sense; everything is then extended to $\mathbb{F}_q(x) = Frac(\mathbb{F}_q[x])$ by multiplicativity of $|\cdot|$.
First note that all elements of $\mathbb{F}_q^\times$ have finite multiplicative order and hence absolute value 1. In other words, the restriction of the value to $\mathbb{F}_q$ is trivial, hence $|\cdot|$ is non-archimedean and we can use ultrametric properties.
Case 1: $|x| \le 1$. Then because $| \cdot |$ is non-archimedean, all elements of $\mathbb{F}_q[x]$ have value $\le 1$. Let $I := \{f \in \mathbb{F}_q[x]: |f| <1\}$. This is an ideal in $\mathbb{F}_q[x]$ (why?).
If $I = \{0\}$, the value is trivial. This is a third case that can occur and seems to have been forgotten by whoever asked you.
So let's assume $I \neq \{0\}$, and let $f$ be the monic polynomial such that $I$ is generated by $f$ (in other words, the monic polynomial of least degree in $I$). It is irreducible (why?). Finally notice that if
$g = u \cdot f^n$
with $u$ a polynomial not divisible by $f$ (in other words, $u \in \mathbb{F}_q[x] \setminus  I$), then $|g| = |f^n| = |f|^n$. With $c := \log_{q_f}(|f|)$, we have $|g| = |g|_{f(x)}^c$ for all polynomials $g$.
(However, the definition of $|g|_{f(x)}$ should be $q_f^{-\nu_f(g)}$ with $\nu_f(g)$ the highest power of $f$ that divides $g$. The way you wrote it in the question, without the "minus" in the exponent, it is not an absolute value.)
Case 2: $|x| > 1$. Then because $1 < |x| < |x|^2 < |x|^3 < \cdots$, and the value defines an ultrametric, and is trivial on $\mathbb{F}_q$, we have for any polynomial $g = \sum_{i=0}^{\deg(g)} a_i x^i \in \mathbb{F}_q[x]$:
$|g| = \max_{i=0}^{\deg(g)} |a_i x^i| = \max_{i: a_i \neq 0} |a_i x^i| = |x^{\deg(g)}| = |x|^{\deg(g)} $.
And as you already wrote, with $c := \log_q(|x|)$, we thus have $|g|=|g|_\infty^c$.
