How is the order of statements gives a different answer?

Question

arent both statements equivalent? so both should be false?

Answer 1

No. Both statements are not equivalent. In the first one, it says there is an $x$ ( a real number) for each and every real number $y$ such that $xy=1$. This is clearly untrue. For example, for $y=5$, the only $x$ that will yield $xy=1$ will be $\frac15$. But this $x=\frac15$ is not the multiplicative inverse of any other real number.

In the second statement, it says for every $y$ there exists an $x$ such that $xy=1$. This is different in that it says that the $x$ for each $y$ does not have to be the same real number as opposed to what was implied by the first statement. And, of course, this is true. Because, for any $y=z$, the multiplicative inverse would be $\frac1z$.

Answer 2

No, what you showed is correct

$$\exists x \in \mathbb{R}^\ast: \forall y \in \mathbb{R}^\ast: xy=1\tag{i}$$

is false; there is no $x$ that works for all $y$, but

$$\forall y \in \mathbb{R}^\ast: \exists x \in \mathbb{R}^\ast: xy=1\tag{ii}$$

is true, as for every $y$ we can find an $x$.

This is how the semantics of these statements is defined. Read up on it.. There are hidden brackets here, really:

$$\exists x \in \mathbb{R}^\ast: \left(\forall y \in \mathbb{R}^\ast: xy=1\right)\tag{i'}$$

e.g. and the $x$ is a "constant" (bound variable) within the bracketed statement. So the statement holds if we can find an $x$ such that for that specific $x$ the inner statement holds, and for that, for each $y$ the statement $xy=1$ should hold, which it does not. Etc.