The “path across the sky” on a given day is close to due east-west, modified by a slight drift north or south, caused by the angle of the moon’s orbit to the equator and it’s position in its orbit-- i.e. how far north or south the moon appears to move during the course of a given day. This will always be a very small fraction of a degree.
However, I suspect you may be more interested in the analemma formed by its precession at “high moon” (by analogy to “high noon”) during the month
IIRC, the Earth orbits at a angle of 25° 35’ to the solar ecliptic (or from a geocentric view, the sun revolves around the earth at that angle to the Earth’s equator). This orbital tilt determines the “height” of the analemma (“figure 8”) progression of the position of the sun at solar noon over the course of the year. The width is determined by the variation in orbital velocity as the the earth moves from apogee (furthest) to perigee (closest approach) in its slightly eccentric elliptical orbit.
The latitude of the observer is the other major factor determining the shape of this figure-8. Essentially, the further you are from the equator, the more lopsided the “8” becomes (one loop becomes much larger than the other)
To a naive observer on Earth, the “apparent” (geocentric) physics is much the same for the Moon orbits at a 5° 8’ angle to the equator, creating a much shorter figure 8, and the eccentricity of the lunar orbit (and hence its variation in orbital velocity) is greater, resulting in a “fatter” figure-8. Latitude plays the same role as it does withthe sun. The fact that the Moon is much closer than the sun also makes these (and other) variations more apparent.
There is a complicating factor, however. The Earth’s solar orbit takes 365.24 days so sun does not return to quite the same spot every year, but the leftover .24 day drift (by far the largest of many complicating factors) is small compared to a year, and can be ignored in casual use. The moon’s orbit (seen from the perspective of an Earth observer, aka the sidereal month) is 27.32 days, and the leftover .32 days is a much larger fraction of the orbital period. The “drift” from orbit to orbit is therefore much larger.
The combined effects of greater orbital eccentricity and drift also explain why nearly 60% of the moon’s surface is visible from the Earth at various times, even though the Moon is tidally locked to Earth.
That’s a general description of the moon’s precession at “high moon” across the sky. I sense you are looking for something more precise, but I hope this is enough to help you at least plot a reasonable first approximation, and that flaws in my recollections don’t embarrass me to much.