Solution:

$ f:[-1,1] \rightarrow R \\ $ is given as,

$$ f(x)=\frac{x}{(x+2)} \\ $$

Let, $$ f(x)=f(y) \\ $$

$$ \begin{aligned} &\Rightarrow \frac{x}{x+2}=\frac{y}{y+2} \\\\ &\Rightarrow x y+2 x=x y+2 y \\\\ &\Rightarrow 2 x=2 y \\\\ &\Rightarrow x=y \\ \end{aligned} $$

$\therefore f$, is a one-one function.

It is clear that $f:[-1,1] \rightarrow$ Range f is onto.

$\therefore f:[-1,1] \rightarrow$ Range f is one-one and onto and therefore, the inverse of the function:

$f:[-1,1] \rightarrow$ Range f exists.

Let g : Range $f \rightarrow[-1,1]$ be the inverse of f.

Let y be an arbitrary element of range f.

Since, $ f:[-1,1] \rightarrow \\ $ Range f is onto, we have:

$$ \begin{aligned} &y=f(x) \text { for same } x \\\\ &\Rightarrow y=\frac{x}{x+2} \\\\ &\Rightarrow x y+2 y=x \\\\ &\Rightarrow x(1-y)=2 y \\\\ &\Rightarrow x=\frac{2 y}{1-y}, y \neq 1 \\ \end{aligned} $$

Now,

let us define $g$ : Range $f \rightarrow[-1,1]$ as $$ g(y)=\frac{2 y}{1-y}, y \neq 1 . \ $$

Now,

$$ (g \circ f)(x)=g(f(x))=g\left(\frac{x}{x+2}\right)=\frac{2\left(\frac{x}{x+2}\right)}{1-\frac{x}{x+2}}=\frac{2 x}{x+2-x}=\frac{2 x}{2}=x \\ $$

$$ (f \circ g)(y)=f(g(y))=f\left(\frac{2 y}{1-y}\right)=\frac{\frac{2 y}{1-y}}{\frac{2 y}{1-y}+2}=\frac{2 y}{2 y+2-2 y}=\frac{2 y}{2}=y \\ $$

$$ \begin{aligned} &\therefore g \circ f=[-1,1] \text { and fog }=\mathrm{I}_{\text {Rangef }} \\\\ &\therefore f^{-1}=g \\ \end{aligned} $$

$$ \Rightarrow \quad f^{-1}(y)=\frac{2 y}{1-y}, y \neq 1 \\ $$