How to convert DFA to regex with State Elimination
If we're talking about the same Elimination problem then you should add an accept state and a start state. In addition, I allowed myself to disregard state $q_b$ as it never part of the language.

The start state will replace $q_0$. The accept state will replace $q_9$. Then to your request, we'll do the rest, starting from $q_1$.

From $q_1$ you either have 0 endless times or a 1 in other words, $0^* \cup1$ from $q_1$ to $q_2$.

Let us now rip $q_3$: from $q_2$ to $q_4$ you'll need 01. While from $q_4$ to $q_1$ you'll need 00.

Proceeding to rip $\{q_0, q_2\}$, note this is no more than a name and could just as well be renamed to $q_5$. Thus from $q_2$ to $q_1$ we have: $11*0$

All that's left is to close the circles which are probably cause your trouble:
For the fun of it, we'll go by order. Let us rip $q_1$:
This means $q_0$ reaches $q_2$ using a concatenation of 0 with the resulting regex of $q_1$ to $q_2$ which is we found and was $0^*1$. Removing $q_1$ then completes the self-cycle of $q_2$ and there would now be an arrow to self: $11^*0$.

Nearly done, remove $q_2$: remove $q_2$ thus connecting $q_0$ to $q_4$ through $0((0^*1)11^*0)^*01$, however, there are also the loops that are created to be taken into account, see the drawing attached: I realized that this is very confusing to read so I've added the following drawing: 



I might have missed a character or two, but those are my two cents, also- remember to keep the loops in mind and their final transitions.

P.s. I noticed in the second NFA (yes, NFA as there are $\epsilon$ transitions), I have accidentally forgotten to add the transition itself from $q_2$ to $q_1$, however it is there in the next NFA