What are the magnitude and direction...

Point charge 3.0 microcoulombs is located at x = 0 m, y = 0.30 m, point charge -3.0 microcoulombs is located at x = 0 m, y = -0.30 m. What are the magnitude and direction of the total electric force that these charges exert on a third point charge Q = 5.0 microcoulombs at x = 0.40 m, y = 0 m?

I already found the net force that acts on the third charge - it's .65 N. the second part of the problem where it's asking for direction is what i am stuck on. i need to find the angle theta CLOCKWISE from the +x (positive x) direction.

No calculation is needed for the direction. By symmetry, the x-components of the 2 forces on Q will cancel and the y-components will result in a force in the -y direction. So the angle is 90 degrees clockwise from the +x direction.

If this is not clear:
call the +3.0 microcoulomb charge q1 and
call the - 3.0 microcoulomb charge q2.

q1 exerts a repulsive force,F, on Q. There is a component (Fx) in the +x direction and a component (Fy) in the -y direction.

q2 exerts an attractive on Q. The force is the same magnitude as F (because |q1|= |q2| and because the distance to Q is the same). The position of q2 is the reflection of the position of q1 in the x-axis. This means the components of the force on Q are -Fx and Fy.

Fx and -Fx cancel, and the 2 Fy's add to give a resultant force in the -y direction of magnitude 2Fy.
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For information. once you've done the above you can calculate the resultant force easily as follows:

Using Pythagoras, the distance from q1 (or q2) to Q is 0.50m

The magnitude of each of the 2 forces = k.q1.Q/d^2 = 8.99 x 3x10^-6 x 5x10^-6/0.5^2
= 0.5394N

If we take the line joining q1 to Q, it makes an angle A, to the y-axis. A = acos(3/5)

The total force = 2Fy = 2 x 0.5394 x cos(A)
= 2 x 0.5394 x 3/5
= 0.65N (in agreement with your value)