How to negate the satement "If $3y$ is a perfect cube then there exists an even number $n$ so that $y = 3n^3$" 
Let $x \in \mathbb{Z}$. If $3y$ is a perfect cube then there exists an even number $n$ so that $y = 3n^3$.

Question: I am a little bit confused here. I know this statement is false and I would like to find 2 things.
(1) What is the negation of this statement? Important: If I want to disprove a statement, do I need to take the negation of the WHOLE implication like $\neg (a \implies b)$? Or just take the negation by parts like $a \implies b \equiv \neg a \implies \neg b$?
I need to find a counter example, but according to a professor I need to find it using the negation.
(2) In what cases would you use $\neg(a \implies b)$ versus $\neg a \implies \neg b$? 


(1) What is the negation of this statement? Important: If I want to disprove a statement, do I need to take the negation of the WHOLE implication like $\neg (a \implies b)$? Or just take the negation by parts like $a \implies b \equiv \neg a \implies \neg b$?

This is important, yes! You ALWAYS take the negation of the whole statement, not the parts. The negation of an implication is NOT the negation of its parts. What we do is
$$
\lnot (a \implies b) = a \land (\lnot b).
$$
In the end you should get the statement: "$3y$ is a perfect cube and for all even numbers $n$, $y \ne 3n^3$."

(2) In what cases would you use $\neg(a \implies b)$ versus $\neg a \implies \neg b$? 

Always the first one. Never the second, when negating a statement.
$\lnot a \implies \lnot b$ may come up if you are doing a proof and you want to prove the converse of a statement $a \implies b$. But don't worry about that.
