Since the foci are F1(-3,0) and F2(3,0), we conclude that

    - the major axis is horizontal y= 0 and coincides with x-axis;

    - the center is at (0,0), half way between the foci;

    - the distance between the foci is 3 - (-3) = 6 units and the distance
      from the center to the foci c = 6/2 = 3 units.


So, the general equation of the hyperbola is

     -  = 1,

where a > 0 is the horizontal transfer semi-axis.



Now consider this condition "for any point on it, the absolute value of the difference of its distances from the foci is 3.".

We know that 'a' is the distance along x-axis from the center (0,0) to any of the two vertices of the hyperbola. 


Then the points (-a,0) and (a,0) are the points on this hyperbola.


For point (a,0), the distance from F1 is (3-a); the distance from F2 is (a+3).
The difference of these distance must be 3, according to the condition.
So, we write this equation

    (a+3) - (3-a) = 3,

     2a = 3,

      a = 3/2 = 1.5.


So, the vetices are at  (-1.5,0) and (1.5,0),  and 'a', as we found, is 1.5.


Then from equation a^2 + b^2 = c^2 we can find 'b^2'

    1.5^2 + b^2 = 3^2

    b^2 = 3^2 - 1.5^2 = 9 - 2.25 = 6.75.


Hence, the standard equation of this hyperbola is

     -  = 1.    <<<---===  ANSWER