Solution:

It is given that, $ f: \mathbf{R} \rightarrow\{x \in \mathbf{R}:-1\lt x \lt1\} \\ $

defined by, $ f(x)=\frac{x}{1+|x|}, x \in \mathbf{R} \\ $

Suppose f(x) = f(y), where x, y ∈ R.

$$ \Rightarrow \frac{x}{1+|x|}=\frac{y}{1+|y|} \\ $$

It can be observed that if x is positive and y is negative, then we have:

$$ \frac{x}{1+x}=\frac{y}{1-y} \Rightarrow 2 x y=x-y \\ $$

Since x is positive and y is negative:

$$ x \gt y \Rightarrow x-y \gt0 \\ $$

But, $2 x y$ is negative.

Then, $2 x y \neq x-y$.

Thus, the case of x being positive and y being negative can be ruled out.

Under a similar argument, x being negative and y being positive can also be ruled $\therefore x$ and y have to be either positive or negative.

When x and y are both positive, we have:

$$ f(x)=f(y) \Rightarrow \frac{x}{1+x}=\frac{y}{1+y} \Rightarrow x+x y=y+x y \Rightarrow x=y \\ $$

When x and y are both negative, we have:

$$ f(x)=f(y) \Rightarrow \frac{x}{1-x}=\frac{y}{1-y} \Rightarrow x-x y=y-y x \Rightarrow x=y \\ $$

$\therefore f is one-one. Now, let y \in \mathbf{R}$ such that $-1\lt y \lt1$.

$$ \begin{aligned} &x=\frac{y}{1+y} \in \mathbf{R} \\\\ &\frac{y}{1+y}=\frac{y}{1+y-y}=y \\ \end{aligned} $$

If y is negative, then there exists,

$$ x=\frac{y}{1-y} \in \mathbf{R} \text { such that } \\ $$

If y is positive, then there exists,

$$ f(x)=f\left(\frac{y}{1-y}\right)=\frac{\left(\frac{y}{1-y}\right)}{1+\left|\left(\frac{y}{1-y}\right)\right|}=\frac{\frac{y}{1-y}}{1+\frac{y}{1-y}}=\frac{y}{1-y+y}=y \\ $$

$\therefore f$ is onto.

Hence, f is one-one and onto.