Explain why the following proof of the conjecture $\exists x \in \mathbb R\forall y \in \mathbb R (xy^2 = y -x )$ is incorrect I'm asked to explain why the proof of the following theorem is incorrect.

Incorrect theorem: 
$\exists x \in  \mathbb R\forall y \in \mathbb R (xy^2 = y -x )$
Proof: 
Let $x = \frac{y}{y^2 +1}$, then $y - x = y - \frac{y}{y^2 +1} = \frac{y^3}{y^2 +1} = \frac{y}{y^2 +1} \cdot y^2 = xy^2$

My attempt:
Theorem asserts that there is some $x$ for which $xy^2 = y -x $ holds no matter what value of $y$ is. Thus to prove the theorem, we need to define $x$ in terms of some constant. In other words, proof should go like this:

Suppose x = $a$ (where $a$ is some real number)
....Proof of the $xy^2 = y - x $ goes here....
Therefore, we showed that if $x=a$ then $xy^2 = y - x $

However, in the proof above $x$ has been defined in terms of $y$, where $y$ is variable. In other words, value of $x$ will change as $y$ changes. Therefore, proof didn't show that $\exists x \in  \mathbb R\forall y \in \mathbb R (xy^2 = y -x )$
Is the explanation correct?
 A: The proof you're given proves the wrong statement. It proves $\forall y \in \mathbb R: \exists x \in \mathbb R:~xy^2=y-x$, or in other words "For all real $y$ there exists a real $x$ satisfying that equation $xy^2=y-x$."
But the statement you're given is $\exists x \in \mathbb R: \forall y \in \mathbb R:~xy^2=y-x$, or in other words "There exists a real number $x$, such that for every single $y$ that equation $xy^2=y-x$ is true."
Those two statements are very different statements. One is true, as that proof proved, the other is false, as that proof should have proved.
Indeed, the proof of one statement can't be used to prove the other statement, because it's wrong. For any $x$ there's at least some $y$ such that $xy^2 \neq y-x$.
A: Yes, your explanation is correct. The reason that the proof is incorrect is that the value of $x$ is stipulated before the value of $y$ is, so $x$ cannot be taken to depend on $y$. If you have to prove that the theorem is incorrect, you can try to prove the negation of the theorem is correct. The negation would be: $\forall x \in \mathbb{R}, \exists y \in \mathbb{R} \hspace{0.25em} \text{such that} \hspace{0.25em} xy^{2} \not = y - x$. Given an $x$, just choose $y = 1$ (other values work too), and you have $x \not = 1 - x$, which is true.
