Answer
$f^{-1}(x)=1-\sqrt{x+1},\quad x\geq-1$
The domain of $f^{-1 }$ is $[-1,\infty)$
The range is $(-\infty,1].$
Work Step by Step
Solve for x. Add 1 to each side and recognize a perfect square.
$y+1=x^{2}-2x+1 ,\quad x\leq 1$
$y+1=(x-1)^{2} ,\quad x\leq 1$
$-\sqrt{y+1}=x-1 ,\quad x\leq 1$
(when $x\leq 1$, the RHS is $\leq 0$, so the LHS should be too)
$x=1-\sqrt{y+1}$
Replace $x $with $f^{-1}(x)$
$f^{-1}(x)=1-\sqrt{x+1},\quad x\geq-1$
The domain of $f^{-1 }$ is $[-1,\infty)$;
the range equals the domain of f, $(-\infty,1].$
$\begin{align*}
f(f^{-1}(x))&=[f^{-1}(x)]^{2}-2f^{-1}(x) \\
&=(1-\sqrt{x+1})^{2}-2(1-\sqrt{x+1}) \\
&=1-2\sqrt{x+1}+x+1-2+2\sqrt{x+1} \\
&=x \\ \\
f^{-1}(f(x))&=1-\sqrt{f(x)+1} \\
&=1-\sqrt{x^{2}-2x+1}, \quad x\leq 1 \\
&=1-\sqrt{(x-1)^{2}}, \quad x\leq 1 \\
&=1-|x-1|, \quad x\leq 1 \\
&=1-[-(x-1)], \quad x\leq 1 \\
&=1+x-1 \\
&=x \end{align*} $
