How would I go about in proving this in Fitch?
$\operatorname{Dodec}(e)$
$\operatorname{Small}(e)$
$\neg\operatorname{Dodec}(e) \lor \operatorname{Dodec}(f) \lor \operatorname{Small}(e)$
Goal: $\operatorname{Dodec}(f).$
I am relatively new to this with no prior experience hence why I am stuck with such question.
With my thinking, I think we would need two subproofs with the conclusion of each being dodec(f) and use disjunction elimination to conclude dodec(f).
1[This is what I have at the moment]
As I pointed out in my comment above, the goal $\ \text{Dodec(f)}\ $ is not provable from the premises \begin{align} &\text{Dodec(e)}\\ &\text{Small(e)}\ &\text{and}\\ &\neg\text{Dodec(e)}\vee\text{Dodec(f)}\vee\text{Small(e)}\ , \end{align} because if a proposition $\ q\ $ is true, so is the proposition $\ \neg p \vee r\vee q\ $, even if $\ r\ $ is false. Therefore, if $\ p\ $, $\ q\ $ and $\ r\ $ are atomic propositions, then $\ r\ $ cannot be a consequence of $\ p\ $, $\ q\ $ and $\ \neg p \vee r\vee q\ $.
In your attempted proof you have already obtained $$\ \text{Dodec(e)}\implies\text{Dodec(f)}\ $$ and $$\ \neg\text{Small(e)}\implies\text{Dodec(f)}\ ,$$ given the premises. You can also easily obtain $$\ \text{Dodec(f)}\implies\text{Dodec(f)}\ .$$ Thus, if you were to replace your third premise with $$ \neg\text{Dodec(e)}\vee\text{Dodec(f)}\vee\neg\text{Small(e)}\ , $$ you would then be able to conclude $\ \text{Dodec(f)}\ $ by disjunction elimination.