Can't determine if given relation is equivalence relation Definition of relation  ~    

$(a,b)$ ~ $(c,d)$ $\iff$ $bc^2=da^2$,
where $(a,b),(c,d)$ are from $\mathbb{R}\times\mathbb{R}$ and $(a,b),(c,d)$ are different from $(0,0)$ 

First of all, I wonder if R can be equivalence relation if it is not defined on $\mathbb{R} \times \mathbb{R}$($(0,0)$ is not included ), because I will not be able to find classes that divide the whole $\mathbb{R} \times \mathbb{R}$.

If I can observe $\mathbb{R} \times \mathbb{R}$ without $(0,0)$, then I should see if this relation is reflexive, symmetric and transitive.
Clearly, this relation is reflexive, but I have troubles with determining if it is symmetric and transitive. 
( Because of elements like $(a,0)$ or $(0,b)$ and dividing with zero )
Also, if it is, I can't imagine what classes are.

Anything you can think of will be helpful.
Thank you in advance.
Reflexive: 
We need to prove that $(a,b)R(a,b)$, let us check if this is true. We have $$ba^2 = ba^2 \implies (a,b)R(a,b).$$

So our relation is reflexive.

Symmetric: We need to prove that if we have $(a,b)R(c,d)$ then this implies $(c,d)R(a,b)$. We have $$bc^2 = da^2 \implies da^2 = bc^2 \equiv (c,d)R(a,b).$$

So our relation is symmetric. 

Transitive: We need to show that if we have $(a,b)R(c,d)$ and $(c,d)R(e,f)$ then we have $(a,b)R(e,f)$. We do so below: We need to show $be^2 = da^2$

$$bc^2 = da^2 \text{ and } de^2 = fc^2 \implies c^2 = \frac{de^2}{f}$$

Substituting the latter into the first equation, we get $$b \cdot \frac{de^2}{f}  = da^2 \iff be^2 = fa^2 \equiv (a,b)R(e,f).$$ 

To complete the proof of transitivity, you need to consider the case $f=0$, look in the comment section for how to do so. 

Since our relation is reflexive, symmetric and transitive, it must be an equivalence relation and as you can see, there was no need to divide by $0$ or any other funky stuff. You can split the set $\mathbb{R} \times \mathbb{R}$ not including $(0,0)$ into equivalence classes.

Note: I used $R$ instead of ~ for the relation.
Unless otherwise stated, assume that all $x$-coordinates are nonzero. Also note that this answer is mostly for fun; I wanted to visualize the equivalence classes!

If we say that $(x_1, y_1) \sim (x_2, y_2)$ if and only if $y_1x_2^2 =  y_2x_1^2$, then (assuming neither $x_1$ nor $x_2$ is $0$) dividing both sides by $x_1^2x_2^2$ we have

$$\frac{y_1}{x_1^2} = \frac{y_2}{x_2^2}.$$

Now, choose a fixed $x_1$ and $y_1$, and let $k =\frac{y_1}{x_1^2}$. Any pair $(x, y)$ related to $(x_1, y_1)$ satisfies $\frac{y}{x^2} = k = \frac{y_1}{x_1^2}$, so that $y = kx^2$. In other words, pairs $(x_1, y_1)$ and $(x_2, y_2)$ with nonzero $x$-coordinates are related if their $y$-coordinates are proportional to the square of their $x$-coordinates, with the same constant of proportionality. 

For a fixed constant of proportionality $k$, we have then that points satisfying the equation $y = kx^2$ are related. For points with nonzero $x$-coordinates, that the relation is an equivalence relation follows from the properties of equality (namely, that equality is an equivalence relation). 

Note that all points whose only nonzero coordinate is $y$ are related (as they correspond to $k = 0$). I'll let you think about what to do with points whose $x$-coordinate is $0$, and what all they're related to.

Thus, (most of) the equivalence classes can be visualized as the graphs of parabolas $$y = kx^2,$$ with one equivalence class for each choice of a real number $k$. Since each of these parabolas go through $(0, 0)$, it should be clear why we needed to remove it from the set under consideration: Every other pair would have been related to it!