Is a point inside regular hexagon

If you reduce the problem down to checking {x = 0, y = 0, d = 1} in a single quadrant, you could make very simple.

public boolean IsInsideHexagon(float x0, float y0, float d, float x, float y) {
    float dx = Math.abs(x - x0)/d;
    float dy = Math.abs(y - y0)/d;
    float a = 0.25 * Math.sqrt(3.0);
    return (dy <= a) && (a*dx + 0.25*dy <= 0.5*a);
}

• dy <= a checks that the point is below the horizontal edge.
• a*dx + 0.25*dy <= 0.5*a checks that the point is to the left of the sloped right edge.

For {x0 = 0, y0 = 0, d = 1}, the corner points would be (±0.25, ±0.43) and (±0.5, 0.0).

You can use the equations for each of the sides of the hexagon; with them you can find out if a given point is in the same half-plane as the center of the hexagon.

For example, the top-right side has the equation:

-sqrt(3)x - y + sqrt(3)/2 = 0

You plug in this the coordinates of the point and then the coordinates of the center. If the results have the same sign, then the point is in the bottom-left half-plane (so it may be inside the hexagon).

You then repeat by using the equations of the others sides.
Note that this algorithm will work for any convex polygon.