We can do this using calculus.
Coordinatize by placing the center of the circle, of radius $r$, at the origin. Generalizing slightly, I'll take $\angle COE = \theta$ instead of specifically $\pi/3$; thus,
$$C = r(0,\sec\theta) \qquad D = r(-\sin\theta,\cos\theta) \qquad E = r(\sin\theta,\cos\theta)$$
As OP notes, the centroid of region $CDFE$ lies on $\overline{CM}$, so its $x$-coordinate is $0$. One sees that the $y$-coordinate of that centroid must match that of the half-region $CFE$, which is bounded by $\overleftrightarrow{CE}$ ($f(x) = - x \tan\theta + r \sec\theta$) and the circle ($g(x) = \sqrt{r^2-x^2}$).
By the formula for the centroid of a bounded region,
$$\begin{align}
\bar{y} \cdot (\text{area}\;CFE) &= \frac12\int_{0}^{r\sin\theta}f(x)^2 - g(x)^2 \;dx \tag{1a}\\[4pt]
&= \frac12\int_{0}^{r\sin\theta}( x^2\tan^2\theta - 2 r x\tan\theta\sec\theta + r^2\sec^2\theta) - (r^2-x^2) \;dx \tag{1b}\\[4pt]
&= \frac1{2\cos^2\theta}\int_{0}^{r\sin\theta} x^2\sin^2\theta - 2 r x\sin\theta + r^2 - r^2\cos^2\theta + x^2\cos^2\theta) \;dx \tag{1b}\\[4pt]
&= \frac1{2\cos^2\theta}\int_{0}^{r\sin\theta} x^2 - 2 r x\sin\theta + r^2\sin^2\theta \;dx \tag{1c}\\[4pt]
&= \frac1{2\cos^2\theta}\int_{0}^{r\sin\theta} \left( x - r\sin\theta\right)^2 \;dx \tag{1d}\\[4pt]
&= \left.\frac1{6\cos^2\theta} \left( x - r\sin\theta \right)^3\;\right|_{0}^{r\sin\theta} \tag{1e}\\[4pt]
&= \frac{r^3\sin^3\theta}{6\cos^2\theta} \tag{1f}
\end{align}$$
(Note: We could get from $(1a)$ to $(1d)$ fairly immediately by observing that $f(x)^2-g(x)^2$ gives the "power", with respect to the circle, of a variable point along $\overline{CE}$. But I digress ...) Then, since
$$\begin{align}
\text{area}\;CFE &= \text{area of }\; \triangle COE - \text{area of sector}\;FOE \tag{2a}\\[4pt]
&= \frac12 \cdot r \cdot r\tan\theta - \frac12 r^2 \cdot \theta \tag{2b}\\[4pt]
&= \frac12 r^2 (\tan\theta - \theta) \tag{2c}
\end{align}$$
we have
$$\bar{y} = \frac{r\sin^3\theta}{3\cos\theta(\sin\theta-\theta \cos\theta)} \qquad\stackrel{\theta=\pi/3}{\to}\qquad
\frac{3r\sqrt{3}}{2(3\sqrt{3} -\pi)} = r\cdot 1.26454\ldots
\tag{$\star$}$$
Alternatively, we can use geometric decomposition.
Writing $\bar{p}$ for the $y$-coordinate of the centroid of $\triangle DCE$ and $\bar{q}$ for the $y$-coordinate of the centroid of sector $DFE$, we have
$$\bar{y} \cdot(\text{area} \;CDFE) = \bar{p}\cdot (\text{area}\; \triangle DCE) - \bar{q}\cdot (\text{area}\; DFE) \tag{3}$$
We "know" that a triangle's centroid is $1/3$ of the way up along a median, and its area is $1/2$-base-times-height, so
$$\begin{align}
\bar{p} \cdot (\text{area}\;DCE) &= \left( r\cos\theta + \frac13 r ( \sec\theta - \cos\theta ) \right) \cdot \frac12 \cdot 2r\sin\theta \cdot r(\sec\theta - \cos\theta) \tag{4a}\\
&= \frac{r \sin^3\theta}{3 \cos^2\theta} \left( 1 + 2 \cos^2\theta\right) \tag{4b}
\end{align}$$
Consulting a convenient list of centroids, we find
$$\bar{q}\cdot(\text{area}\;DFE) = \frac{4 r \sin^3 \theta}{3(2\theta - \sin 2\theta)}\cdot \frac{r^2}{2}(2\theta-\sin 2\theta) = \frac23 r^3 \sin^3 \theta \tag{5}$$
So, the right-hand side of $(3)$ is $(4b)-(5)$, which reduces to twice the value of $(1f)$. Since the area of $CDFE$ is likewise twice the value in $(2c)$, the "twice"s cancel, and $(3)$ yields the result shown in $(\star)$. $\square$
