First check that if you approach (0, 0) along x = 0 or y = 0, then x^4 < y < x^2 is never the case, so the limit is 0.
If the line through (0, 0) is not vertical or horizontal, then it is described by an equation y = kx for some k <> 0. Therefore, f(x, y) = 1 on this line iff x^4 < kx < x^2.
There are several cases to consider here. Assume that x > 0. Then we can divide by x without changing the inequality: x^3 < k < x. So, iff . If k >= 1, then this is never the case, so approaching (0, 0) along such line from the right has limit 0. If k < 1, then such limit is also 0 because eventually x <= k. However, as k becomes smaller and smaller, you have f(x, y(x)) = 1 for x that are arbitrarily close to 0. This means that in every neighborhood around (0, 0) there are points (x, y) such that f(x,y) = 0 and other points (x', y') such that f(x',y') = 1.
There may be some cases left to consider.