Proving $2x^2-3xy+y^2=0$ is transitive and anti-symmetric or symmetric and reflexive. Let $R$ be the binary relation defined on $\mathbb{R}$ by $xRy$ iff $2x^2-3xy+y^2=0$
For reflexive we get $2x^2=2x^2\implies-x=x$ which means reflexive on $xRx$ 
$2x^2-3xy+y^2=0$ tried going for $2y^2-yz+z^2=0$ then adding them together but now I'm stuck with a long useless equation any tips of proving this transitive as for anti symmetric i know $2y^2-3yx+x^2=0$ in case $x=y$ so it should be anti symmetric but I don't know how to say it.

$\text{R}$ is a binary relation on $\mathbb{R}$ defined as $x\text{R}y\iff 2x^2-3xy+y^2=0$
Reflexive:
If $x=y$, then $2x^2-3xy+y^2=2x^2-3x^2+x^2= 0$
Hence, we have $x\text{R}x\ \ \forall \ x \in\mathbb{R}$ and the relation is Reflexive

Now using a technique called Completing the Square we have:
$$\begin{align}x\text{R}y &\iff 2x^2-3xy+y^2=0\\
&\iff x^2-\frac{3}{2}xy+\frac{y^2}{2}=0\\
&\iff x^2-\frac{3}{2}xy+\frac{y^2}{2}+\big(\frac{3y}{4}\big)^2-\big(\frac{3y}{4}\big)^2=0\\
&\iff\big(x-\frac{3y}{4}\big)^2-\big(\frac{y}{4}\big)^2=0\\
&\iff \big(x-\frac{y}{2}\big)\big(x-y\big)=0\\
&\iff x=\frac{y}{2} \ \ \text{Or} \ \ x=y \ \ \ \ \ \ \ \ \ \ \ \ -(1)
\end{align}$$

Antisymmetric:
We will proceed by contradiction. Let us assume $\text{R}$ is not antisymmetric. Then $\exists \ x,y\in\mathbb{R}$ such that $x\neq y$ and both $x\text{R}y$, $y\text{R}x$ are satisfied. Since $x\neq y$, by $(1)$ we have the following $$x\text{R}y\Rightarrow x=\frac{y}{2}$$ $$y\text{R}x\Rightarrow y=\frac{x}{2}$$
Thus,
$$x= y=0$$
Which leads to a contradiction. Hence, $\text{R}$ is antisymmetric.

Transitive:
The relation is not transitive because $3\text{R}6$ and $6\text{R}12$ but $3\text{R}12$ does not hold.

Symmetric:
The relation is not symmetric because $6\text{R}12$ but $12\text{R}6$ does not hold.
