Given continuous $f_n(x)\ge 0$ are there continuous $g_n(x)\ge 0$ so that $f_n(x)g_n(x)\to +\infty$ whenever all $f_n(x)\neq 0$? Let $f_n:X\to [0,+\infty)$ be continuous functions, where $X$ is a metric space. Then do there exist continuous $g_n:X\to [0,+\infty)$ so that $f_ng_n\to +\infty$ poinwise in $\cap_n[X\setminus f_n^{-1}(\left\{0\right\})]$?

I am particularly interested in the special case where if $f_n(x)=0$ for some $n$ and $x$ then $f_m(x)=0$ for all $m\ge n$. As a special case of the special case, do such $g_n$ exist when  $f_n=f_{n-1}b_n$ for continuous $b_n\ge 0$?
Let's try for $f_ng_n\simeq n$. At the zero set of $f_n$ this of course can not be satisfied, so one has to modify this equality slightly. Multiply with $f_n$ to get a positive term 
$$
f_n^2g_n\simeq nf_n
$$
and then desingularize
$$
(n^{-2}+f_n^2)g_n=nf_n
$$
So taking 
$$
g_n=\frac{n^3f_n}{1+n^2f_n^2}
$$
gives a function sequence that has a chance to come close to your requirements. But as said in the other answer, without further assumption on $f_n$ it will be hard to prove pointwise divergence close to the zero set.
Not without some additional assumption on the $f_n$'s.  Consider the sequence $f_n = 0$ for all n.