Proof verification: Determine whether $f(x)$ is one-to-one or not 
$$f:\mathbb{R}\mapsto\mathbb{R}
\text{ defined as } f(x)=2x^3+3x-4 \\
\underline{\textit{proof(by contradiction)}}\\
\text{By definition a function is called one-to-one if}\\
f(x)=f(y),\text{ } x,y\in\mathbb{R} \text{ implies that x=y.}\\ 
\text{Suppose for the sake of contradiction }f(x)=f(y)\text{ but }x\neq y.\\ \text{Therefore } 2x^3+3x-4 = 2y^3+3y-4\\
2x^3-2y^3+3x-3y=0\\
2(x-y)(x^2+xy+y^2)+3(x-y)=0\\
(x-y)\left [2(x^2+xy+y^2)+3)\right ]=0\\
\text{which implies that either } x=y\text{ or }2(x^2+xy+y^2)+3=0 \\
\text{which contradicts to the assumption that } x\neq y \Rightarrow \Leftarrow  $$


Is there anything missing in my proof or there is a better solution? I got confused at the either or part because $x=y$ my or may not be true. My logic is given above, if there is something I misunderstand please give your feedback.

$\textbf{FINAL EDIT}$
  Here is the continuation of the proof that it is never true that the right factor $x^2+xy+y^2=-\frac{3}{2}$. 
  $$\text{We know that } x^2\geq0 \text{ and } y^2\geq0\Rightarrow x^2+y^2\geq 0 \text{ and that there are two cases for }x,y.\\
\textbf{case 1: }\text{If nonzero x and y have the same sign then we are done since that implies that }\\
 x^2+xy+y^2>0\\
\textbf{case 2: } \text{If nonzero x and y have differing signs }\\ xy<0 
\text{ and also we know that } (x+y)^2>0\\
 \Rightarrow x^2+2xy+y^2>0 \\
 \Rightarrow x^2+xy+y^2>-xy\\
\text{ Since } xy<0  \Rightarrow -xy>0 \Rightarrow x^2+xy+y^2>0  \\
\text{Therefore it is never true that } x^2+xy+y^2=-\frac{3}{2}$$


Is my proof complete now?

In your edit, Case 1 is fine, but the "proof" in Case 2 isn't. If the assumption of Case 2 is that precisely one of $x$ and $y$ is negative, then you have to add Case 3 to you proof: the case when both $x<0$ and $y<0$. If the assumption of Case 2 is that at least one of $x$ and $y$ is negative, then it also includes the possibility that both $x<0$ and $y<0$, which makes your claim that $xy<0$ invalid. (Not to mention that $x$ and $y$ can also be equal to $0$, so you have to either include that in existing cases or consider even more cases.) In any event, the next issue is: where did the claim that $x^2+xy+y^2\ge0$ come from?
A much shorter proof can be achieved by completing the squares, for example like this:
$$x^2+xy+y^2=x^2+2\cdot x\cdot\frac{1}{2}y+\frac{1}{4}y^2-\frac{1}{4}y^2+y^2=\left(x+\frac{1}{2}y\right)^2+\frac{3}{4}y^2\ge0.$$
Then $2(x^2+xy+y^2)+3\ge\ldots?$
