So you want to prove the following theorem:
Theorem: If $\Gamma,\phi \vdash \psi$ and $\Gamma, \phi \vdash \neg \psi$, then $\Gamma \vdash \neg \phi$
Proof:
First, I'll assume that you can use the Deduction Theorem, which states that for any $\Gamma$, $\varphi$, and $\psi$:
If $\Gamma \cup \{ \varphi \} \vdash \psi$, then $\Gamma \vdash \varphi \rightarrow \psi$
So if $\Gamma,\phi \vdash \psi$ and $\Gamma, \phi \vdash \neg \psi$, then by the Deduction Theorem we have $\Gamma \vdash \phi \to \psi$ and $\Gamma \vdash \phi \to \neg \psi$
This means that if can show that $\phi \to \psi, \phi \to \neg \psi \vdash \neg \phi$, then we're there.
This is not easy, but here goes:
First, let's prove: $\phi \to \psi, \psi \to \chi, \phi \vdash \chi$:
\begin{array}{lll}
1&\phi \to \psi & Premise\\
2& \psi \to \chi & Premise\\
3&\phi& Premise\\
4&\psi& MP \ 1,3\\
5&\chi& MP \ 2,4\\
\end{array}
By the Deduction Theorem, this gives us Hypothetical Syllogism (HS): $\phi \to \psi, \psi \to \chi \vdash \phi \to \chi$
Now let's prove the general principle that $\neg \phi \vdash (\phi \to \psi)$:
\begin{array}{lll}
1. &\neg \phi& Premise\\
2. &\neg \phi \to (\neg \psi \to \neg \phi)& Axiom \ 1\\
3. &\neg \psi \to \neg \phi& MP \ 1,2\\
4. &(\neg \psi \to \neg \phi) \to (\phi \to \psi)& Axiom \ 3\\
5. &\phi \to \psi& MP \ 3,4\\
\end{array}
With the Deduction Theorem, this means $\vdash \neg \phi \to (\phi \to \psi)$ (Duns Scotus Law)
Let's use Duns Scotus to show that $\neg \phi \to \phi \vdash \phi$
\begin{array}{lll}
1. &\neg \phi \to \phi& Premise\\
2. &\neg \phi \to (\phi \to \neg (\neg \phi \to \phi))& Duns \ Scotus\\
3. &(\neg \phi \to (\phi \to \neg (\neg \phi \to \phi))) \to ((\neg \phi \to \phi) \to (\neg \phi \to \neg (\neg \phi \to \phi)))& Axiom \ 2\\
4. &(\neg \phi \to \phi) \to (\neg \phi \to \neg (\neg \phi \to \phi))& MP \ 2,3\\
5. &\neg \phi \to \neg (\neg \phi \to \phi)& MP \ 1,4\\
6. &(\neg \phi \to \neg (\neg \phi \to \phi)) \to ((\neg \phi \to \phi) \to \phi)& Axiom \ 3\\
7. &(\neg \phi \to \phi) \to \phi& MP \ 5,6\\
8. &\phi& MP \ 1,7\\
\end{array}
By the Deduction Theorem, this means $\vdash (\neg \phi \to \phi) \to \phi$ (Law of Clavius)
Using Duns Scotus and the Law of Clavius, we can now show that $ \neg \neg \phi \vdash \phi$:
\begin{array}{lll}
1. &\neg \neg \phi& Premise\\
2. &\neg \neg \phi \to (\neg \phi \to \phi)& Duns \ Scotus\\
3. &\neg \phi \to \phi& MP \ 1,2\\
4. &(\neg \phi \to \phi) \to \phi& Clavius\\
5. &\phi& MP \ 3,4\\
\end{array}
By the Deduction Theorem, this also means that $\vdash \neg \neg \phi \to \phi$ (DN Elim or DNE)
Finally, we can show the desired $\phi \to \psi, \phi \to \neg \psi \vdash \neg \phi$:
\begin{array}{lll}
1. &\phi \to \psi& Premise\\
2. &\phi \to \neg \psi& Premise\\
3. &\neg \neg \phi \to \phi& DNE\\
4. &\neg \neg \phi \to \psi& HS \ 1,3\\
5. &\neg \neg \phi \to \neg \psi& HS \ 2,3\\
6. &(\neg \neg \phi \to \neg \psi) \to (\psi \to \neg \phi)& Axiom \ 3\\
7. &\psi \to \neg \phi& MP \ 5,6\\
8. &\neg \neg \phi \to \neg \phi& HS \ 4,7\\
9. &(\neg \neg \phi \to \neg \phi) \to \neg \phi& Clavius\\
10. &\neg \phi& MP \ 8,9\\
\end{array}
Now, you can actually get a little more quickly to $\neg \neg \phi \vdash \phi$ as follows:
\begin{array}{lll}
1&\neg \neg \phi&Premise\\
2&\neg \neg \phi \to (\neg \neg \neg \neg \phi \to \neg \neg \phi)&Axiom \ 1\\
3&\neg \neg \neg \neg \phi \to \neg \neg \phi&MP \ 1,2\\
4&(\neg \neg \neg \neg \phi \to \neg \neg \phi) \to (\neg \phi \to \neg \neg \neg \phi) & Axiom \ 3\\
5& \neg \phi \to \neg \neg \neg \phi & MP \ 3,4\\
6&(\neg \phi \to \neg \neg \neg \phi) \to (\neg \neg \phi \to \phi) & Axiom \ 3\\
7& \neg \neg \phi \to \phi & MP \ 5,6\\
8&\phi&MP \ 1,7\\
\end{array}
However, since the proof of $\phi \to \psi, \phi \to \neg \psi \vdash \neg \phi$ relies on Clavius, I took the road that I did.
Best Answer
In a proof by reductio ad absurdum, you begin by assuming the negation of the proposition to be proven; in this case, you would begin by assuming $$\neg\Bigl([\neg D \lor (A \land B)] \longrightarrow[(J \rightarrow \neg A) \rightarrow (D \rightarrow \neg J)]\Bigr).$$ The objective is to deduce an absurdity (something of the form $P\land \neg P$). As you can see, you don't do either of those (you neither start with the negation of the proposition to be proven, nor do you deduce an absurdity), so you did not give a proof by reductio ad absurdum.
In a "conditional proof" of $P\to Q$, you assume $P$, and then you proceed to deduce $Q$. From this proof, thanks to the Deduction Metatheorem, you can conclude that there is a proof of $P\to Q$. In the case at hand, your proof would have to begin by assuming $$\neg D\lor (A\land B)$$ and finish by deducing $$(J\to \neg A)\to(D\to\neg J).$$ Again, as you can see, you did not do this.
So I think that "what is wrong" with your proof is that it did not follow the instructions given; you were asked to produce a proof by one of two methods, and you did not. That is not to say your proof is invalid: from what I can tell, it is a valid proof. You provided a conditional proof of the contrapositive of the proposition you want to establish. Unfortunately, a conditional proof of the contrapositive is neither a conditional proof, nor a reductio ad absurdum proof. As to partial marks, that's up to your instructor.
Now, don't despair: it is fairly easy to turn your proof into a proof by reductio ad absurdum. Start with the negation of the proposition you have; you will be able to deduce from this assumption your line 1. Having deduced your line 1, you will end up, with your development, deducing $\neg(\neg D\lor (A\land B))$. Now, form the original assumption you should also be able to deduce $(\neg D\lor (A\land B))$, and now you are in the situation you wish: you will have $P\land \neg P$, with $P\equiv (\neg D\lor (A\land B))$.