Proceed methodically: Suppose the premisses are true and conclusion false. So
$1.\quad p \quad \Rightarrow \quad T$
$2.\quad p \lor q \quad \Rightarrow \quad T$
$3.\quad q \to (r \to s) \quad \Rightarrow \quad T$
$4.\quad t \to r \quad \Rightarrow \quad T$
$5.\quad \neg s \to \neg t \quad \Rightarrow \quad F$
From the last, you know
$6.\quad \neg s \quad \Rightarrow \quad T$
$7.\quad \neg t \quad \Rightarrow \quad F$
Whence
$8.\quad s \quad \Rightarrow \quad F$
$9.\quad t \quad \Rightarrow \quad T$
4 and 9 give us
$10.\quad r \quad \Rightarrow \quad T$
So $r \to s$ is false, and hence, from (3)
$11.\quad q \quad \Rightarrow \quad F$.
So we've worked backwards to successfully find a valuation of all the variables (at lines 1, 8--11) which you can check makes all of 1 to 5 true, i.e. makes the premisses true and conclusion false.
Systematizing this "working backwards" method gives us the user-friendly method of "semantic tableaux" or "truth-trees" used in many textbooks (including mine, and Paul Teller's which is freely available online).
$(1) + (4)$ do not imply $q\lor r$, so undo line $(5)$ We do have that $(3) + (4)$ imply $p$. If $p \lor q$, and $\lnot q$, then $p$. Perhaps this can be line $(5)$.
Suggestion for the next step $(6)$: from $(3),$ along with the premise $\lnot q \rightarrow (u \land s)$, it follows by modus ponens that $u \land s$.
(7) Now, extract $s$ from $u \land s$.
(8) Then introduce the conjunction: $p \land s$. $p$ is from your new (5th) step, and $s$ from step (7).
Now we can use the premise $p\land s \implies t$ and $p \land s$ from step (8) to conclude by modus ponens that $t$, as desired
Key point: If you haven't used a premise, think of how it might help get you from what you have to what you need to establish. Always keep the goal or target proposition in your mind.
Best Answer
Joshua Taylor has given a correct proof using one common set of inference rules, but it appears, from the work you've shown, that your system of inference rules is a little different. So let me try it with rules that look like what you used. I'll start with the conclusion, $(p\to q)\to(p\land\neg r)$ that you arrived at. The "implication rule" that you used to go from 2 to 3 should allow me to convert this to $(\neg(p\to q))\lor(p\land\neg r)$. Next, a distributive law (admittedly dual to the one you used going from 7 to 8, so I hope both versions of distributivity are available to you) gives me $[(\neg(p\to q))\lor p]\land[(\neg(p\to q))\lor\neg r]$. Then conjunctive simplification, which got you from 1 to 4, should produce $(\neg(p\to q))\lor\neg r$. Finally, the implication rule, now in the other direction, as in your inference from 8 to 9, gives $(p\to q)\to\neg r$, as desired.