I agree that the original statement is false for the integers. However, the formula you wrote is true for the integers! Indeed, given any x, take y = 1. Then the premise of (x < 0 ∧ y < 0) -> x < y is false, so the whole statement is true.

It helps to think in terms of proof of formulas. What should a proof of the statement "Every negative number has another negative number greater than it" do? Given any x together with an evidence that x is negative, it mustproducesome y and two pieces of evidence: that y is negative and that x < y. Note that producing the evidence that y < 0 is the proof's responsibility; it is not given to it for free in the same way as the evidence that x < 0 is given. In general, a proof of ∀x P(x) is given x while a proof of ∃x P(x) has to produce x. Similarly, a proof of P -> Q is given an evidence for P and it has to produce an evidence for Q. For this problem, this means that y < 0 should be on the right-hand side of the implication: ∀x∃y[x < 0 -> (y < 0 ∧ x < y)]. In fact, I prefer ∀x[x < 0 -> ∃y(y < 0 ∧ x < y)] because here the proof can use the evidence that x < 0 to help it come up with y.

Constructing negation is very simple. First, you put ¬ in front of the whole formula. The problem says to move ¬ inside the quantifiers, which you can do using the following equivalences:

¬∀x P(x) <=> ∃x ¬P(x)

¬∃x P(x) <=> ∀x ¬P(x)

If you'd like, you can move ¬ further inside using

¬(P -> Q) <=> P ∧ ¬Q

¬(P ∧ Q) <=> ¬P ∨ ¬Q

¬(P ∨ Q) <=> ¬P ∧ ¬Q

¬¬P <=> P

The original statement is true on rationals because for each positive rational x there exists a smaller positive rational, e.g., x/2.