The way I approached this initially was to work out the quadratic that fits the given points in f(x) and the cubic that fits the given points in g(x).
f(x)=-4x²/5+11x/5+4, g(x)=-x³/6-2x²/3+31x/6+8/3.
f(x)=(-4x²+11x+20)/5, g(x)=(-x³-4x²+31x+16)/6.
f(x)/g(x)=(6/5)(-4x²+11x+20)/(-x³-4x²+31x+16).
However, there may be other curves fitting the given points, so we should examine how else we can look at the problem. The set of points for g is larger than that for f, and there are only two points having the same x coordinate in each set. This reduces f(x)={ (-1,1), (4,0) } and g(x)={ (-1,-3), (4,2) }.
f(x) and g(x) are effectively y values, so f/g is a comparison of y values, that is, y coordinates for common x coordinates.
From this f/g={ -⅓, 0 }. We can’t compare f and g for the other points because they do not share x coordinates. The domain for f/g is { -1, 4 }, whereas the domain for f is { -1, 0, 4 } and for g is { -1, 1, 4, 5 }.