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.
I'll give a summary "sketch" of one proof approach.
I'd suggest you start from the first premise:
First check what follows from $r\land \lnot s$ and then what follows when $q \land \lnot s$. (At least one of these must be true. Why?)
From each of the above, we can use conjunction elimination (or simplification) to obtain $\lnot s$. And so, we can use disjunction elimination to conclude $\lnot s$.
From $\lnot s$ and premise $(2)$, we can derive (infer), by modus ponens, $(p\land r)\rightarrow u$.
From $(p \land r) \rightarrow u$ together with premise $(3)$, we get $(p\land r) \rightarrow (s \land \lnot t)$
Now, we already derived $\lnot s.$ This means $\lnot s \lor t$ is true. Why?
But $\lnot s \lor t \equiv \lnot (s \land \lnot t).$
From $\lnot (s \land \lnot t),$ together with our derived $(p\land r )\rightarrow (s\land \lnot t)$, by modus ponens, we get $\not(p \land r)$
This means that either $\lnot p$, or $\lnot r$ is true.
If $\lnot p$ is true, we can build from this (addition) to get $\lnot p \lor q \equiv p\rightarrow q$, as desired.
If $\lnot r$ is true, then $r\land \lnot s$ is false (first alternative from premise $(1)$. Then it must follow that $q\land \lnot s$ is true. So it follows that $q$ is true (simplification). And given $q$, we have $\lnot p \lor q$ (addition), which is equivalent to $p \rightarrow q$, as desired.
Therefore, from $\lnot (p \land r)\equiv \lnot p \lor \lnot r$ we can conclude $(q\rightarrow q)$.
I'll leave this to you to write formally (including keeping proper track of temporary assumptions), with proper justifications.
Best Answer
You can think about it in this way: $(A\to B)\land(\neg A\to C)$ is a rule you're given, which tells you what happens when you know the truth value of $A$. If $A$ is true, then you must have $B$; if $A$ is false, you must have $C$. And since $A\lor\neg A$ is a tautology (always true), then either you have $A$ and $B$ or you have $\neg A$ and $C$, which we can write as $(A\land B)\lor(\neg\land C)$.
More formally: \begin{align}(A\to B)\land(\neg A\to C)&\equiv (\neg A\lor B)\land(A\lor C)\\&\equiv(\neg A\land A)\lor(\neg A\land C)\lor(B\land A)\lor(B\land C)\\&\equiv(\neg A\land C)\lor(B\land A)\lor(B\land C)\\&\equiv(\neg A\land C)\lor(B\land A)\end{align}
Some details on the above derivation:
Finally, a deduction: