Step 1: A Calculus Based Route:
A function that is strictly increasing (or strictly decreasing) on its whole domain must be one-one, because it never repeats a value.
We show f is strictly increasing on R by checking its derivative on each piece separately.
Step 2: Derivative for x Greater Than or Equal to 0:
For $x\ge0$, $f(x)=\dfrac{2x}{1+x}$.
Differentiating using the quotient rule gives
\[ f'(x)=\dfrac{2(1+x)-2x(1)}{(1+x)^2}=\dfrac{2}{(1+x)^2} \]
This is positive for every $x\ge0$, so f is strictly increasing here.
Step 3: Derivative for x Less Than 0:
For $x<0$, $f(x)=\dfrac{2x}{1-x}$.
Differentiating gives
\[ f'(x)=\dfrac{2(1-x)-2x(-1)}{(1-x)^2}=\dfrac{2}{(1-x)^2} \]
This is also positive for every $x<0$, so f is strictly increasing here too.
Step 4: Combine the Two Pieces:
f is continuous at $x=0$ since both pieces give $f(0)=0$.
Since f is strictly increasing on $x<0$, strictly increasing on $x\ge0$, and continuous at the join point, f is strictly increasing on the whole of R.
A strictly increasing function never takes the same value twice, so $f(x_1)=f(x_2)$ forces $x_1=x_2$.
Final Answer:
Since f is strictly increasing on all of R, it is one-one, hence proved.
\[ \boxed{f \text{ is one-one}} \]