Answer:
(3, -5.4)
Explanation:
You want the center of mass of a plane region of uniform density bounded by the parabola y = 3x -x² and the line y = -3x.
Bounds
The height of the region is the difference between the boundaries:
h(x) = (3x -x²) - (-3x) = 6x -x² = x(6 -x)
The height is zero at x=0 and x=6, where the factors of h(x) are zero.
Symmetry
The axis of symmetry of this height function is halfway between the zeros, at x = (0+6)/2 = 3.
Area
The area of the region is the integral of the height function between the bounds where h(x) = 0:

Y-dimension
The y-coordinate of the center of mass is the integral of the product of a differential of mass and its y-dimension, divided by the area.

Dividing this moment by the area, we find the desired y-coordinate to be ...

X-dimension
The x-coordinate of the center of mass is on the line of symmetry, at ...

The center of mass is (3, -5.4).
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Additional comments
Alternatively, you could take advantage of the published coordinates of the centroid of a parabolic region: 4/10 of the height above the flat edge, and on the line of symmetry. Here, the region is bounded by the line at x=3:
y = -3x = -3(3) = -9
and the parabola at x=3:
y = x(3 -x) = 3(3 -3) = 0
Then the y-coordinate of the centroid is ...
-9 +0.4(0 -(-9)) = -9 +3.6 = -5.4 . . . . . as above
centroid = (3, -5.4)
In the integral to find the y moment, we used the centroid of the differential area as the average of the upper and lower bounds of that area:
((3x -x²) + (-3x))/2 = -x²/2
Modern calculators do this integration easily, achieving full calculator precision.