Broad hints:
You may use analytic geometry. It's easy to show that a centroid has coordinates $(x_1+x_2+x_3)/3$ and $(y_1+y_2+y_3)/3$ $\;$ where $(x_1,y_1)$ and etc. are the coordinates of the triangle vertices. You just need to use the fact that medians intersect at the centroid which divides the medians in ratio $1/3$.
The line segment AB with $A(x_1,y_1)$ and $B(x_2,y_2)\;$ divided by some point on it with ratio $k$ yields the coordinates (the basics of analytic geometry):
$$\frac{x_1+k\,x_2}{1+k}\;;\;\;\frac{y_1+k\,y_2}{1+k}$$
Now it easy to find the coordinates of the centroid, and I'm leaving it out for you giving only the answer here $(x_1+x_2+x_3)/3\;$and $(y_1+y_2+y_3)/3\,.$
Just as easy is to find the centroid of a new triagle with points D, E, F which divides the sides of the original triangle ABC in the same ratio, say, $\;p.\;$ We use the formulas above to find the vertices or our new triangle:
$$\bigg(\frac{x_1+p\,x_2}{1+p}\;;\;\;\frac{y_1+p\, y_2}{1+p}\bigg);\quad \bigg(\frac{x_2+p\,x_3}{1+p}\;;\;\;\frac{y_2+p\, y_3}{1+p}\bigg);\quad \bigg(\frac{x_3+p\,x_1}{1+p}\;;\;\;\frac{y_3+p\, y_1}{1+p}\bigg)$$
Now we find its centroid. Make sure you can derive all the formulas above using the properties of medians and the formulas for the line segment. I'm talking about deriving $(x_1+x_2+x_3)/3.\;$ The rest is easy. It's the same procedure:
$$\bigg(\frac{x_1+p\,x_2}{1+p}+\frac{x_2+p\,x_3}{1+p}+\frac{x_3+p\,x_1}{1+p}\bigg)/3=(x_1+x_2+x_3)/3$$