Propositional Logic: p → q, ¬p → r, ¬q → ¬r ⊢ q Is this proof correct?
p → q, ¬p → r, ¬q → ¬r ⊢ q
1) p → q      premise
2) ¬p → r     premise
3) ¬q → ¬r    premise
  4) -q       assumption
  5) -p       MT 1,4
    6) p      assumption
    7) ⊥      ¬e 5,6
  8) ⊥        ¬e 2,3
9) q          ¬e 8

Am I going correct? What might be the next step?
I updated my answer.
 A: There exist different natural deduction systems with different rules for negation.  For this exercise you might assume the negation of q.  
Notice that ($\lnot$p $\rightarrow$ r) and ($\lnot$q $\rightarrow$ $\lnot$r) have r and $\lnot$r as their right hand side or consequent, respectively.  What do your negation rules say when you have an instance of some well-formed formula $\alpha$ and $\lnot$$\alpha$?  
If you assume $\lnot$q and then assume p, can you infer some other instance of $\alpha$ and $\lnot$$\alpha$?  What was the last assumption made which you might discharge?  If you discharge it, then with the other premisses, can you then discharge the first assumption/hypothesis/supposition?
A: Your step 8 makes no sense. Also, your other steps with the logical structure clear are as follows:

$p \to q$   [premise]
$\neg p \to r$   [premise]
$\neg q \to \neg r$   [premise]
If $\neg q$:
  $\neg p$   [MT]
  If $p$:
    Contradiction.
  [Therefore you get nowhere because you did not get contradiction under the first assumption.]

Note that it is never useful to create a new subcontext by assuming the negation of what you have already proven. In the above, in the context where $\neg q$ you proved $\neg p$, so it would be useless to create the subcontext where $p$, since you already know it can never occur. Instead, from $\neg p$ and $\neg q$ you can derive other things by modus ponens. See if you can complete the proof.
I encourage you to use indentation or some other syntax to make clear to yourself what the context of every sentence is. A sentence may not be true in the global context (no assumptions) but may be true in some context (under some assumptions).
The final proof should look like:

$p \to q$   [premise]
$\neg p \to r$   [premise]
$\neg q \to \neg r$   [premise]
If $\neg q$:
  $\neg p$   [MT]
  ... [Use the premises here to deduce two sentences that contradict one another] ...
  Contradiction.
$\neg \neg q$.
$q$.
A: The proof works up to line 5 as user21820 explained.  The assumption made on your line 6 would not be easy to discharge later. Rather take the derived line 5 and continue with it.
Here is a completed proof using a Fitch-style proof checker which you may use to check this and other proofs. 

On lines 7 to 9, I derived $\neg\neg R$ to use modus tollens on line 10. The proof checker does not have an inference rule for double negative introduction. However, it is easy to derive that result when needed.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
A: 
1) p → q      premise
2) ¬p → r     premise
3) ¬q → ¬r    premise
  4) -q       assumption
  5) -p       MT 1,4
    6) p      assumption
    7) ⊥      ¬e 5,6
  8) ⊥        ¬e 2,3
9) q          ¬e 8


Well, you knew what to do, but were not clear on how to do it.   Lines 4,5,8,9 are correct, but lines 6,7 would not derive the contradiction you require.
You have shown that $\neg q$ implies $\neg p$, and since $\neg p\to r$, then it also implies $r$.   However, the third premise $\neg q\to\neg r$ shows $\neg q$ implies $\neg r$ too.   There is your contradiction.
What assuming $p$ would have done, was allow you to derive $\neg p$ using fundamental rules (rather than the derived rule of modus tolens).   So if you cannot use derived rules, do that thing.
$$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline #2\end{array}}\fitch{~~1.~p \to q\\~~2.~ \neg p \to r\\~~3.~ \neg q \to \neg r}{\fitch{~~4.~\neg q}{~~5.~\neg p\qquad~\mathsf{MT}1~,4\\~~6.~r\qquad\quad{\to}\mathsf E~5,2\\~~7.~\neg r\qquad~{\to}\mathsf E~4,3\\~~8.~\bot\qquad~~\neg\mathsf E~6,7}\\~~9.~\neg\neg q\qquad\quad\neg\mathsf I~4{\mapsto}8\\10.~q\qquad\qquad~\neg\neg\mathsf E~9}\qquad\qquad\lower{4ex}{\fitch{1.~p\to q\\~~2.~ \neg p \to r\\~~3.~ \neg q \to \neg r}{\fitch{~~4.~\neg q}{\fitch{4.1.~p}{4.2.~q\quad~{\to}\mathsf E~4.1,1\\4.3.~\bot\quad\neg\mathsf E~4,4.2}\\~~5.~\neg p\qquad\neg\mathsf I~4.1{\mapsto}4.3\\~~6.~r\qquad\quad{\to}\mathsf E~5,2\\~~7.~\neg r\qquad~{\to}\mathsf E~4,3\\~~8.~\bot\qquad~~\neg\mathsf E~6,7}\\~~9.~\neg\neg q\qquad\quad\neg\mathsf I~4{\mapsto}8\\10.~q\qquad\qquad~\neg\neg\mathsf E~9}}$$
