Finding fault in proof [contains logical quantifiers]

[iopz]

Homework Statement

Consider the following incorrect theorem: ∃x∈ℝ ∀y∈ℝ (xy^2 = y-x)

[Translation (not part of the original problem statement): There is at least an x∈ℝ such that, for every y∈ℝ, (xy^2 = y-x).]

What's wrong with the following proof?

Let x = y(y^2+1), then

y-x=y-y/(y^2+1)=y^3/(y^2+1)=y/(y^2+1) * y^2=xy^2​

Homework Equations

1. (xy^2 = y-x)

2. x = y(y^2+1)

3. y-x=y-y/(y^2+1)=y^3/(y^2+1)=y/(y^2+1) * y^2=xy^2

The Attempt at a Solution

Since the first equation is to be proven and the third equation seem to be correct, i think that the problem lies in the second.
I have transformed the theorem as follow:
[∃x∈ℝ ∀y∈ℝ (xy^2 = y-x)] = [∃x(x∈ℝ∧∀y(y∈ℝ→(xy^2=y-x))]
From this, i thought that since one of the things to prove is that there is at least an actual x that is true for all y, the substitution done in equation 2 is not correct (since x is substituted not with an actual value but a free variable).
But I'm not sure if this is really the reason for why the proof is incorrect. Any help will be appreciated.

 

[willem2]

What's wrong with the following proof?

Let x = y(y^2+1)

​

This is already wrong. x is a single number and can't depend on y​

 

[iopz]

This is already wrong. x is a single number and can't depend on y​

I see, thank you!