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?

$$f(x) = f(y)$$
$$\Rightarrow2x^3 + 3x  - 4 = 2y^3 + 3y  - 4$$
$$\Rightarrow 2(x^3-y^3) + 3(x-y) = 0$$
$$\Rightarrow 2(x-y)(x^2+xy+y^2) + 3(x-y) = 0$$
$$\Rightarrow (x-y)(2x^2 + 2xy + 2y^2 +3) = 0$$
$$\Rightarrow x = y$$
The last step can be justified by saying that the right factor is never $0$. I leave this as an exercise for you. 
An alternative solution:
Note that $f'(x) = 6x^2 + 3 = 3(2x^2  +1) > 0$, such that $f$ is strictly increasing and therefore injective. 
