Is a total order compatible with a partial order? I was given the following multipart problem. 
Part 1: Consider the poset ({2,4,6,9,12,18,27,36,48,60,72},|), with the indicated integers and the divides relation. Find the following, if they exist; if they don’t exist, explain why they don’t.
Maximal and minimal elements? Max: 27, 48, 60, 72. Min: 2,9.
Least upper bound of {2,12}? 12.
Greatest lower bound of {60,72}? 12.
I drew out a Hasse diagram and think I solved those 3 questions correctly, although I would appreciate if someone could confirm.
Then comes the next part: 
Suppose we impose the following total order on the set: 2,4,18,6,27,12,36,9,72,60,48. Is this compatible with the partial ordering of the divides relation? Why or why not?
I just don't even get what this is asking at all. I understand what a poset and total ordering are, but what does is it mean "is it compatible?". The only thing I can think of is that it's not actually a total order. Given the Hasse diagram, the order given doesn't seem to work. Any help would be appreciated, thanks.

This answer basically just confirms what it appears from the comments that you’ve already understood.
Let $\langle P,\le\rangle$ be a partial order, and let $\preceq$ be a linear (total) order on $P$. The order $\preceq$ is compatible with the partial order $\le$ if the following condition is satisfied:

for all $x,y\in P$, if $x\le y$, then $x\preceq y$.

In other words, $\le$ and $\preceq$ never assign opposite orders to any pair of elements of $P$; either they assign the same order, or $\le$ doesn’t assign any order to that pair.
In your problem, $\preceq$ is given by 
$$2\prec4\prec18\prec6\prec27\prec12\prec36\prec9\prec72\prec60\prec48\;.$$
As you noted in the comments, this disagrees with the order $\le$ on the pairs $\{6,18\}$, $\{9,18\}$, and $\{9,27\}$, so it’s not compatible with the partial order $\le$.
