$\mathrm{The}\mathrm{Range}\mathrm{of}f\left(x\right)={\sin }^{-1}\left(\frac{1}{x^2-2x+2}\right)\mathrm{is}$

The correct option is:

$(0,\frac{\pi }{2}]$

To find the range of the function $f\left(x\right)={\sin }^{-1}\left(\frac{1}{x^2-2x+2}\right)$, we first analyze the expression inside the inverse sine function.

Let $g\left(x\right)=x^2-2x+2$.
We can rewrite this quadratic expression by completing the square:
$g\left(x\right)={(x-1)}^2+1$

Since the square of any real number is non-negative, we have:
${(x-1)}^2\ge 0$
Adding 1 to both sides:
$g\left(x\right)\ge 1$

Thus, the range of $g\left(x\right)$ is $[1,\infty )$.

Now, we look at the reciprocal term inside the ${\sin }^{-1}$:
Let $y=\frac{1}{g\left(x\right)}=\frac{1}{{(x-1)}^2+1}$

Since $1\le {(x-1)}^2+1<\infty$, taking the reciprocal reverses the inequality:
$0<\frac{1}{{(x-1)}^2+1}\le 1$
This means the argument $y$ lies in the interval $(0,1]$.

Since the function ${\sin }^{-1}\left(y\right)$ is strictly increasing for $y\in [-1,1]$, we can apply the inverse sine to the bounds of the interval:
As $y\to 0^{+}$, ${\sin }^{-1}\left(y\right)\to 0$.
For the upper bound, when $y=1$, ${\sin }^{-1}\left(1\right)=\frac{\pi }{2}$.

Therefore, the range of the function is:
$(0,\frac{\pi }{2}]$