feedback on my solution regarding eqivalence relations. For all $x, y \in \mathbb{R}$ define that $x \equiv y$ if $x^2 = y^2$. Then $\equiv$ is an equivalence
relation on $\mathbb{R}$, there are infinitely many equivalence classes, one of them consists of
one element and the rest consist of two elements.
Answer:
True. 
Reflectivity: 
To show that $\mathbb{R}$ is reflexive we need to show that 
$\forall x \in \mathbb{R} : x \equiv x$. Let $x \in \mathbb{R}$, $x \equiv x$ if $x^2 = x^2$, which is obvious.
Symmetry:
Let $x,y \in \mathbb{R}$ with $x \equiv y$, so we have $x^2=y^2$ and trivially we get $y^2=x^2$, hence $y \equiv x.$
Transitivity:
Let $x,y,z \in \mathbb{R}$ with $x \equiv y$ and $y \equiv z$. We now show that $x \equiv z$: Since $x^2=y^2$ and $y^2=z^2$ we get by transitivity of equality $x^2=z^2$ and therefore $x \equiv z.$
Is the answer to the question correct? 
And please tell me how I can prove that there are infinitely many equivalence classes one of them consists of
one element and the rest consist of two elements.
If it's not correct then please provide the correct answer.

It is right. And FYI, if you define $x \sim y \iff f(x) = f(y)$, for a given function $f$, it will always wield an equivalence relation. The equivalence classes will be the level surfaces of $f$. I think it's a good exercise to prove this.
