1. We need the domain of the composite function $(f\circ g)(x)=f(g(x))=\ln(g(x)).$

   Since $\ln t$ is defined only for $t>0,$ the domain of $f\circ g$ is the set of all real $x$ for which $g(x)>0.$

2. Given $g(x)=\frac{x^4-2x^3+3x^2-2x+2}{2x^2-2x+1}.$

   So we must check:

   • denominator is never zero,
   • numerator is always positive or determine where it is positive.

3. First, examine the denominator: $2x^2-2x+1=2\left(x-\frac12\right)^2+\frac12.$

   Since $2\left(x-\frac12\right)^2\ge 0,$ we get $2x^2-2x+1>0 \quad \text{for all } x\in\mathbb R.$

   Hence the denominator never vanishes and is always positive.

4. Now examine the numerator: $x^4-2x^3+3x^2-2x+2.$

   Rewrite it as $x^4-2x^3+x^2+2x^2-2x+1+1.$

   Grouping terms, $x^4-2x^3+x^2=x^2(x-1)^2,$ and $2x^2-2x+1=2\left(x-\frac12\right)^2+\frac12>0.$

   So $x^4-2x^3+3x^2-2x+2 = x^2(x-1)^2+(2x^2-2x+1)+1.$

   Therefore, $x^4-2x^3+3x^2-2x+2 = x^2(x-1)^2+2\left(x-\frac12\right)^2+\frac32.$

   Every term on the right is non-negative, and the constant term $\frac32$ is strictly positive. Hence $x^4-2x^3+3x^2-2x+2>0 \quad \text{for all } x\in\mathbb R.$

5. Since both numerator and denominator are positive for all real $x$, $g(x)>0 \quad \text{for all } x\in\mathbb R.$

   Therefore, $\ln(g(x))$ is defined for every real number.

6. Hence the domain of $f\circ g$ is $\mathbb R.$

7. Checking options:

   • A: $(0,\infty)$ — incorrect
   • B: $[1,\infty)$ — incorrect
   • C: $\mathbb R$ — correct
   • D: $[0,\infty)$ — incorrect

Therefore, the correct option is C.