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?
 A: $$f(x) = f(y)$$
$$\Rightarrow2x^3 + 3x  - 4 = 2y^3 + 3y  - 4$$
$$\Rightarrow 2(x^3-y^3) + 3(x-y) = 0$$
$$\Rightarrow 2(x-y)(x^2+xy+y^2) + 3(x-y) = 0$$
$$\Rightarrow (x-y)(2x^2 + 2xy + 2y^2 +3) = 0$$
$$\Rightarrow x = y$$
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. 
A: You have "$f(x) = f(y)$" implies that at least one of "$x=y$" or "$2(x^2 + x y+ y^2)+3 = 0$" is true.  If you can show that the second is never possible for any $x,y \in \mathbb{R}$, you would be done.
Now $x^2 + xy + y^2$ has what minimum value?  Can it ever be as small as $-3/2$?
EDIT
In your case 2, you go off the rails after "$x y < 0 \implies x^2 + xy + y^2 > x y$".  You also have not addressed what happens when $x$ and $y$ don't have signs, that is, when $x=0$ or $y=0$.  Something like ...

When $x = 0$, $2(x^2 + x y+ y^2)+3 = 2(y^2)+3 > 0$.  Likewise, when $y=0$ by symmetry.  Consequently, we need only consider nonzero $x$ and $y$.

Then, continuing on, ...

If nonzero $x$ and $y$ have the same sign, then $xy>0$ and, since $x^2 + y^2 >0$, $x^2 + x y+ y^2 > 0$.
  If nonzero $x$ and $y$ have differing signs, then $xy < 0$ and, ... [what goes here?]

Further Edit
I would make small tweaks to your final edit.

If nonzero $x$ and $y$ have differing signs, then $xy < 0$, giving $-xy > 0$.  Then, since $(x+y)^2$ is also positive, 
  $$  x^2 + xy + y^2 = (x+y)^2 - xy > 0 > -3/2 \text{,}  $$
  contradicting $2(x^2 + xy + y^2)+3 = 0$.  Therefore, $x = y$ and we find $f$ is injective.

We immediately transform "$xy < 0$" into the form we will use, so that the reader knows what form of it they are looking for in the next sentence.  We take this and another positivity result to produce the needed positivity result, separating the expression from $-3/2$, and explicit use of that constant reminds the reader what fact they were looking for.  We observe that we have obtained our needed contradiction, then the conclusion forced by that contradiction, then the conclusion forced by that result, unrolling the nested conditional proofs that we are in.
A: 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 $2(x^2+xy+y^2)+3\ge\ldots?$
