If $f(x)=\frac{x^2-x}{x^2+2 x}$, then $\frac{d}{d x}\left\{f^{-1}(x)\right\}$ is equal to

$\frac{3}{(1-x)^2}$

We have, $f(x)=\frac{x^2-x}{x^2+2 x}$

Clearly, f(x) is not defined at x = 0, -2

So, Domain (f) = R - {-2, 0}

For all x ∈ domain(f), we have

$f(x)=\frac{x^2-x}{x^2+2 x}=\frac{x-1}{x+2}=1-\frac{3}{x+2}$

Now, 

$fof^{-1}(x)=x$

$\Rightarrow f\left(f^{-1}(x)\right)=x$

$\Rightarrow 1-\frac{3}{f^{-1}(x)+2}=x$

$\Rightarrow 1-x=\frac{3}{f^{-1}(x)+2}$

$\Rightarrow f^{-1}(x)=\frac{3}{1-x}-2 \Rightarrow \frac{d}{d x}\left\{f^{-1}(x)\right\}=\frac{3}{(1-x)^2}$