Let $$(p\land q)\to r, d\to p, d\to \lnot r \implies \vDash \lnot q$$
Disproving:
we choose $d=f$. Therefore, $p=f, r=t$. Hence, since $(p\land q)\to r = t$ then it must be that $(p\land q)=t$.
Now, no matter if $q$ is true or false, we got a contradiction. Therefore, $\lnot q$ cannot be implied from this set.
Is my justification right?
No, the last step in the first line is not correct, as I explained in the comment. $x \rightarrow$ True does not mean, that $x=$ True. But the result is correct, the implication is false. There will be no such pitfalls, if you simply make a table with all possibilities.
In this case I would take the four possibilities by independently choosing $d$ and $q$, look for those in which all three implications are valid (three remain), and check if in any of those there is a true $q$. If not, this would immediately prove the statement, since all possibilities are accounted for.
But there is one: $d = p =$ false, $q = r =$ true. $p \land q$ is not valid, so the first statement is true too. Everithing is fulfilled and $q$ is true, so the implication is disproved.