Natural deduction proof dilemma 
Give a natural deduction proof of $A \land (B \rightarrow \neg C)$
from hypotheses $A \land D$ and $A \rightarrow \neg C$.

My attempt so far:

*

*$A \land D$

*$A \rightarrow \neg C$

*A (1, $\land e$)

*$\neg C$ (2,3 $\rightarrow e$)

*D (1, $\land e$)

I'm new to the subject, how should I proceed from here?
 A: Since the conclusion is a conjunction you need to derive the two conjunctives separately so that way you can use the rule $I∧$ to get the conclusion. You have already derivate one of them $A$, now you need $B ⇒ ¬C$. Since the main connective is a conditional you have to use the "practical rule of $I⇒$": "assume the antecedent and try to derive the consequent", This is what the comments highlighted.
Basically, the correct way to proceed is to look at the main connective of the conclusion, this will suggest which rule to use and what are the assumptions to be made.
A: Who cares about $D$?
You need to derive $B\to\neg C$, since you have already derive $A$, so you can use $\wedge$ introduction.
To derive a conditional, use conditional introduction.  So derive $\neg C$ under the assumption of $B$.
Oh, look, you have already derived $\neg C$ without that assumption, so you can indeed do so under that assumption too.
You have actually done most of the work, so to finish up...
$${\begin{array}{|l}~~1.~A \land D\\~~2.~A \rightarrow \neg C\\\hline~~
3.~A\qquad (1, \land e)\\~~~~\begin{array}{|l}~~4.~B\\\hline~~5.~\neg C\quad(2,3, \to e)\end{array}\\~~6.~B\to\neg C\quad (4{-}5,\to i)\\~~7.~A\wedge(B\to\neg C)\quad(3,6,\wedge i)\end{array}\\\blacksquare}$$
