Is $\forall x,\exists y (x \ge y^2 \Rightarrow x>0)$ same as $\forall x (\exists y: x\ge y^2\Rightarrow x > 0)$

The first statement: $\forall x \exists y (x\geq y^2 \Rightarrow x > 0)$
The Second statement: $\forall x (\exists y: x\geq y^2 \Rightarrow x > 0)$

Actually, my question is this: In the first statement, there exists $y$ for every $x$ such that what's in the parenthesis holds but in the second statement, "$\exists y$" is placed inside parenthesis. Does it make any difference to the first statement?

To deal with a more general case, I modified the sentences to

$$ S_1)~~\forall x \exists y [ P(x,y) \to Q(x)] $$ $$ S_2)~~\forall x [ \exists y P(x,y) \to Q(x)] $$

$S_1$ is in what is called the prenex normal form. Converting quantified sentences containing material conditionals, $\to$ or $\Rightarrow$, to sentences in prenex form needs special attention.

You might know that $P \to Q$ is equivalent to $\neg P \lor Q$. Let us apply this to the sentence $S_2$.

$$ \begin{aligned} &\forall x [ \exists y P(x,y) \to Q(x)]\\ \Leftrightarrow~ &\forall x [ \neg \exists y P(x,y) \lor Q(x)]\\ \Leftrightarrow~ &\forall x [ \forall y \neg P(x,y) \lor Q(x)] \quad\text{by DeMorgan's law}\\ \Leftrightarrow~ &\forall x [ \forall y (\neg P(x,y) \lor Q(x))] \quad \text{ because $Q(x)$ is independent of $y$}\\ \Leftrightarrow~ &\forall x \forall y [P(x,y) \to Q(x)]\\ \end{aligned} $$

That is, $S_1$ and $S_2$ are not equivalent due to the quantification over $y$.

Additional comment: As I showed for a more general case that $S_2$ and $\forall x \forall y [P(x,y) \to Q(x)]$, which is different from $S_1$, are logically equivalent, the only thing I can do now is to prove each proposition.

Let us restrict our domain of discourse to real numbers.

Consider $\forall x \exists y [ x \ge y^2 \to x > 0]$. Let $x$ be arbitrary. Now, we choose $ y = 1$. Then, if $x \ge y^2 = 1$, we have $x \ge 1 > 0$. Therefore, the claim is true!
Now, consider $\forall x [\exists y: x \ge y^2 \to x > 0]$. Let $x = 0$. We have to show that if $\exists y: x \ge y^2$ is true, so is $x > 0$. However, we know that $x = 0 \ge 0^2$; hence, there is $y = 0$ such that $x \ge y^2$ holds, but $x > 0$ is false. We found $x = 0$ such that $\exists y: x \ge y^2$ but not $x > 0$. Therefore, the claim is false.