Verify if Propositions hold or not I want to show wether or not these two propositions hold or not.
The first one is that $$\forall x\exists y(xy>0\implies y>0)$$
For this one I noticed that hen $y=0$ it doesn’t hold. But I’m confused wether or not I am done proving or not.

The second one is 
$$\lnot\forall x\exists y(xy\ge  x^2)$$
For this one, I noticed that if you set $x=y$, then you get $y^2\ge y^2$. Once again, I’m not sure how this helps.
I would appreciate any help on these two propositions.
 A: For the first proposition, showing that it doesn't work for $y = 0$ isn't a counterexample that disproves the statement. The qualifiers at the start of the statement, $\forall{x} \exists{y} $, mean that for every $x$, there exists at least one $y$ such that $xy \gt 0 \implies y \gt 0$. A counterexample that properly disproves the proposition would be some $x$ for which there doesn't exist any $y$ such that 
$xy \gt 0 \implies y \gt 0$.
For the second proposition, it's easiest to transform the statement, as suggested by BrianO's comment. We have:
$\neg \forall{x}  \exists{y} (xy \geq x^2) \iff \exists{x} \neg  \exists{y} (xy \geq x^2) \iff \exists{x} \forall{y} \neg(xy \geq x^2) \iff \exists{x} \forall{y} (xy \lt x^2)$
To disprove this, we have to show that for all $x$, there exists some $y$ such that $xy \geq x^2$. 
A: *

*$\forall x\exists y(xy>0\to y>0)$ is true.
(by contradiction) Let's prove that the negation of 1. is impossible. Negating 1. gives
$$
  \exists x \forall y (xy>0 \land y\le 0).\tag{not-1}
$$
Suppose $x$ is such that (not-1) holds. Then $\forall y (xy>0 \land y\le 0)$, so $\forall y (y\le 0)$ — that is, every $y$ is less than $0$, which is false. So (not-1) is false, thus 1. is true.
(direct proof) Note that $p\to q$ is equivalent to $\neg p\lor q$, so 1. is equivalent to
$$
\exists x\forall y(xy\le 0 \lor y>0).
$$
Clearly, for all $y$, it's true that $(y\le 0 \lor y>0)$, or equivalently, $(1y\le 0 \lor y>0)$. So $\forall y(1y\le 0 \lor y>0)$. Now we see that $x=1$ satisfies $\forall y(xy\le 0 \lor y>0)$, so 1. follows.

*$\neg\forall x\exists y(xy\ge x^2)$ is false.
The sentence is equivalent to
$$
\exists x\forall y(xy<x^2).\tag{2'}
$$
Suppose $x$ is such that $\forall y(xy<x^2)$. Then in particular it must be true for $y=x$, so we have $x^2 = xy < x^2$, which is ridiculous. So (2') can't be true, thus the equivalent 2. is false.
