Computing the center of mass of a signal using the Fourier transform
The first thing you do is correct. You can calculate the sum of all the values of a signal evaluating its DTFT at $\omega=0$.

When you try to express the numerator with the DTFT though, you make a pretty clever thing but it is not correct, because you are evaluating the DTFT at $\omega=\ln\left( n^{\frac{1}{j2\pi n}}\right)$, which is not always real-valued. Remember that $\omega\in\mathbb{R}$.

To find how the DTFT of $x(n)$ and the numerator of $c$ are related, you can use the following property:

$$x(n)\xrightarrow{\mathscr{F}}X(\omega) \implies nx(n)\xrightarrow{\mathscr{F}}j\frac{dX(\omega)}{d\omega}$$

Using this property and your approach for the denominator (evaluating the transform at $\omega=0$), you should get the desired result.