Find the equation of the line for two of the medians, and compute the intersection.
The line from $v_1$ to the midpoint of the opposite side can be parametrized by:
$$(x,y,z)=(0,0,0)+t(2,1,3)$$
The line from $v_2$ to the midpoint of the opposite side can be parametrized by:
$$(x,y,z)=(0,1,1)+s(1,-1,0)$$
This gives us the system of equations:
$$x=2t=s\\y=t=1-s\\z=3t=1$$
So we see that $t=\frac{1}{3}$ and $s=\frac{2}{3}$ is the intersection, which gives the point $(\frac{2}{3},\frac{1}{3},1)$ as the centroid.
The idea behind this formula is that you decompose the polygon into triangles and find the centroid of each, then find the centroid of the resulting point masses, with the mass at each point equal to the area of the corresponding triangle (i.e., the weighted average of the triangle centroids).
Assume that the polygon is star-shaped with respect to the origin and that the vertices are consecutively numbered in a counterclockwise direction. The polygon can be decomposed into triangles defined by the origin and successive vertices $\mathbf v_i$ and $\mathbf v_{i+1}$. The area of each of these triangles is $\frac12(x_iy_{i+1}-x_{i+1}y_i)$. If we concentrate all of this “mass” at the centroid of the triangle $\mathbf c_i$, then the centroid of the polygon is given by the usual formula for the center of mass of a set of point masses: $$\mathbf c=\frac1A\sum_{i=0}^{N-1}\frac12(x_iy_{i+1}-x_{i+1}y_i)\mathbf c_i.\tag{1}$$ The centroid of a triangle with vertices $\mathbf u$, $\mathbf v$ and $\mathbf w$ is just $\frac13(\mathbf u+\mathbf v+\mathbf w)$. Substituting this into $(1)$ yields the formula in the question.
With a bit more algebra the same formula can be derived for a polygon that’s star-shaped with respect to an arbitrary point. It’s not difficult to show that this formula also holds for a general non-self intersecting polygon. Some of the triangle areas can be negative in this case, but they end up canceling the “extra” area that you get for other triangles.
Best Answer
If you know the area of the polygons, you can use the weighted average of the containing and internal polygon centroids. Specifically, if the two polygons have areas $A_{out}$ and $A_{in}$, and centroids $\bf r_{out}$ and $\bf r_{in}$, the centroid of the polygon "doughnut" will be: $$\frac{A_{out}{\bf r_{out}} - A_{in}{\bf r_{in}}}{A_{out}-A_{in}}$$