How does the centripetal force vary with the speed of rotation for a constant radius of the path?

F=omega^2*r

F=m*v^2/r

Don't call it "centripetal force".  Calling it such makes you think that there is a certain type of force which exists solely to be the "centripetal force".

Centripetal force and centrifugal force are the two most unnecessary and confusing topics in introductory physics, and should be avoided as terms at all costs.

Call it "net force needed to cause centripetal acceleration".  I will be calling it Fnet for short.

It depends on what you mean by "speed of rotation".
1. Do you mean omega, the angular velocity (perhaps in radians per second)?
2. Or do you mean v the linear velocity of rotation?

If case 1, then...
Your formula is
Fnet = m*omega^2*r

A) if you increase omega, but hold both m and r constant, Fnet will increase with the square of omega.

B) if you increase r, but hold both m and omega constant, Fnet will increase with direct proportionality.

If case 2, then...
Your formula is
Fnet = m*v^2/r

A) if you increase v, but hold both m and r constant, Fnet will increase with the square of v.

B) if you increase r, but hold both m and v constant, Fnet will decrease with an inverse-proportional trend with the r.

increase velocity = an increase in the required centripetal force
increase in radius = ?decrease in centripetal force? (not sure on that one)

a = v^2 / r

(a)  With a fixed radius, the faster the rotation, the larger the force.
(b)  With a fixed velocity, the shorter the radius, the larger the force.