Introduction: It turns out from comments that OP's function is actually
and the question asks for the critical point(s), and for a classification.
Note first of all that $x^2-2x+2=(x-1)^2+1$. In particular, $x^2-2x+2$ is always positive.
Note also that the derivative of $\ln(x^2-2x+2)$ is
The derivative is $0$ precisely if $2x-2=0$, that is, if $x=1$. Depending on the local definition of critical point, that means that the critical point is $1$, or that it is $(1,0)$.
Classification: Here we come to the motivation for my answering the question. Distressingly, answers to this and similar questions by the OP work with the second derivative. We can more simply use the first derivative.
As remarked earlier, the denominator in Expression (1) is always positive. Thus (1) is positive precisely if $2x-2$ is positive, and negative precisely if $2x-2$ is negative.
Note that $2x-2$ is negative if $x\lt 1$, and $2x-2$ is positive if $x\gt 1$.
It follows that our original function $\ln(x^2-2x+2)$ is decreasing in the interval $(-\infty,1]$, and increasing in the interval $[1,\infty)$. Thus $\ln(x^2-2x+2)$ attains a local (and absolute) minimum at $x=1$.
Remark: The second derivative test tends to be popular with students, since it promises a mechanical approach to the classification of critical points. Unfortunately, computation of the second derivative can be messy, and therefore subject to error. Often, the manipulation needed to find the critical points is enough to give us the sign of the first derivative in the intervals of interest. This is enough for the required classification.