If a graph G is not bipartite then there exists a preference list assignment to the vertices of G such that no stable matching exists with respect to this assignment.
This is a variant of the Stable marriage problem in the sense that now the graph is not bipartite, so a stable matching may not exist according to the statement, while in the case of bipartite graph we can always do so thanks to the Gale-Shapley algorithm. How do we even approach this problem? I have no idea how to do so. What properties of non-bipartite graph can we use? Any help will be appreciated.
Well, my approach would be to try to see what can happen with small non-bipartite graphs. It is easy to see that if $G$ is a triangle, then there will always be a stable matching, and the same holds if $G$ is a triangle plus a vertex of degree one. Here, I made the adventurous leap to try $G = K_4$, the complete graph on four vertices. It is not difficult to find an example in this case. Let $a,b,c,d$ denote the vertices of $G$ and consider a preference list where $a$ is the least preferred option for any of $b,c$ and $d$, and the most preferred option for $b,c,d$ is, respectively, $c,d,b$. Then you can convince yourself that anyone matched with $a$ (the "smelly roommate") will create an instability with his/her second most preferred option.
This assumes that in this setup of the problem everyone prefers being matched to being unmatched (i.e. that only maximal matchings are considered stable). This is called the stable roommates problem, and there is a tricky algorithm by Irving that decides whether a stable matching exists in this case and finds one if it does, all in linear time in the number of edges (if I remember correctly).