Find the quotient space of $\equiv \subseteq \mathbb R^2$ such that $x \equiv y \iff (\exists t >0)(tx=y)$ 
Find the quotient space of $\equiv \subseteq \mathbb R^2$ such that $x
> \equiv y \iff (\exists t >0)(tx=y)$

The quotient space is the set of all equivalence classes of given equivalence relation, and so the first thing to do is to find those classes. I can think of three classes that when combined cover the whole set of real numbers. Namely $[0]_{\equiv}, [1]_{\equiv}, [-1]_{\equiv}$ And so, the quotient space appears to be: 
$$ \frac{\mathbb R}{\equiv} = \{[0]_{\equiv}, [1]_{\equiv}, [-1]_{\equiv} \}$$
Where 
$[0]_{\equiv} = \{0 \}$ 
$[1]_{\equiv} = \{y \mid y>0 \}$ 
$[-1]_{\equiv} = \{y \mid y<0\}$ 
Is it the correct solution?
 A: The problem asks to find a nice set of representatives for each element of the quotient space, so then we can identify this set of representatives with the quotient space. Hence the quotient space is "going to be" this set of representatives.
Now it appears that you are not computing the requested quotient space but an analogous quotient for $\mathbb{R}$. Your solution for this different problem is correct though.
One answer for the original problem, as hinted by Masacroso, is giving by the set of representatives $X=\mathbb{S^1}\cup \{0\}=\{\vec x\in \mathbb{R}^2 \mid \  \|\vec x\|=1 \}\cup \{\vec 0\}$.
Indeed, any element of $\mathbb{R}^2/\equiv$ is equivalent to an element in $X$ because $[x]=[\frac x {\|x\|}]$ with $\frac x {\|x\|}\in X$. And also two different elements in $X$ are not equivalent because if $x,y\in X$ and $x\equiv y$ then we have $x=t y$ with $t>0$ and hence $1=\|x\|=\|ty\|=t\|y\|=t$ so $x=ty=y$.
Nevertheless, this is not the unique solution to the problem because there are other nice set of representatives that could be though as the solution. For example the set of representatives $$Y=\{(x,y) \mid y=-1\} \cup {(0,0)} \cup \{(1,0)\}$$
is other possible solution to the problem. 
In some sense the first solution is better than the second because the "shape" of this set is compatible with the equivalence relation in some sense. (This can be made precise with the notion of Quotient topology, but probably you should not worry about this for the moment)
If you have any questions, do not hesitate to ask in the comments.
