Fake proof of one-to-one function

Question

Prove or disprove that $f: \mathbb{R} \rightarrow \mathbb{R}$ is one-to-one if,

$$f(x) = -3x^2+7$$

Assume $f(x) = f(y)$, then

$$-3x^2+7 = -3y^2+7$$

$$-3x^2 = -3y^2$$

$$x^2 = y^2$$

$$x = y$$

Hence function is one-to-one.

While we can easily see that function is not one-to-one. For example $f(-1)=f(1)$.

Where is the proof wrong?

Answer 1

The proof is wrong because $$x^2=y^2\iff x=\pm y$$ and not just $x=y$.

Answer 2

$$x^2 = y^2 \Leftrightarrow x = \pm y$$