By how much does Haumea's fast rotation affect its surface gravity? The equation for surface gravity is $\frac{GM}{r^2}$ but I'm not sure how to include the effects from its rotation.
The numbers are going to be approximate because Haumea's size is uncertain.  It may have a ring.  If it does, then it's smaller, if it doesn't, then it's larger.
The formula is simple enough.   For force:    $F= \frac {m v^2} {r} $ where v is the rotational velocity and r is the same r in the gravitational formula.   or $\frac {v^2} {r}$ if you want the reverse acceleration.
Haumea has a rotational period of 3.915 hours or 14,090 seconds (I'm rounding) and an average radius of 780,000 meters (assuming it has a ring, I'm going to go with that assumption), but an estimated equatorial radius of about 1,050,000 meters and that's the number we need.  Divide the larger radius by number of seconds, multiply by 2 Pi and the velocity is 468.2 m/s.   A tiny bit faster than Earth's equatorial velocity.
Some Maths, the lifting force on the equator is $ \frac {468.2^2} {1,050,000} $ = $ 0.208 \frac {m}{s^2} $, slightly over half it's listed equatorial surface gravity of $ 0.401 \frac {m}{s^2} $.
That's a pretty high ratio.   Possibly the highest in our solarsystem for objects that are massive enough to be spheroid shaped by their mass and gravity.   There are asteroids that are thought to effectively have negative gravity at their equator, held together by the stickiness of the material, but those are a lot smaller.