I was wondering if someone could tell me if my answer & working out is correct as I don't have access to the answer. I am to find the negation of $\exists x \forall y((y > x) \land (x = 3y)$ : $$ = \forall x \neg \forall y((y > x) \land (x = 3y)) \\ \iff \forall x \neg \forall y((y > x) \land (x = 3y)) \\ \iff \forall x \exists y \neg ((y > x) \land (x = 3y)) \\ \iff \forall x \exists y(\neg(y > x) \vee \neg(x = 3y)) \\ \iff \forall x \exists y((y \leqslant x) \vee (x \neq 3y)) $$
Is $$\forall x \exists y((y \leqslant x) \vee (x \neq 3y))$$ negation of $$\exists x \forall y((y > x) \land (x = 3y))$$ ?
If I am incorrect please do tell me where I went wrong, thank you in advance.
As Andre said, your answer is roughly correct except for two things:
"$t \ne u$" denotes a short-hand for "$\neg( t = u )$" if the formal system does not natively support the symbol "$\ne$". This may seem silly, but many textbooks and references on logic define formal systems with equality but without the unequal symbol. Why? Simply because we don't need that and it makes it harder to prove facts about the formal system itself. Nevertheless, in actual mathematical practice it is far simpler to have and use "$\ne$".
"$t \le u$" normally denotes a short-hand for "$t < u \lor t = u$", and "$t > u$" is a short-hand for "$u < t$". If this is the case for your formal system, then it is not obvious that "$\neg( t > u )$" is equivalent to "$t \le u$". In fact, without further assumptions on the ordering, they may not be equivalent. If the domain is intended to be the real numbers, we do assume transitivity and irreflexivity and trichotomy (which are expressed by "$\forall x,y,z\ ( x < y \land y < z \to x < z )$" and "$\forall x\ ( \neg( x < x ) )$" and "$\forall x,y\ ( x < y \lor y < x \lor x = y )$"), in which case we can prove (try it!) that "$\neg( t > u )$" is equivalent to "$t \le u$" (the forward implication needs trichotomy while the backward implication needs transitivity and irreflexivity).