Answer:

>! $$cos~x + cos~y$$

Solution: 

>! Approximate the x and y intercepts of the loop as $$\frac{\pi}{3}$$ (since $$2^{-\frac{8}{5}} \simeq \frac{1}{3.03}$$). Then we need $$h(\pm \frac{\pi}{3}, \pm \frac{\pi}{3}) = 1$$, so to turn those $$\pm\frac{\pi}{3}$$s into something reasonable we probably need $$cos~x$$ (there are other ways, of course, but it was hinted that we need trig and also it takes care of both positive and negative very compactly: $$cos~\frac{\pi}{3} = cos~\frac{-\pi}{3} = \frac{1}{2}$$). It's likely to be symmetric in $$x$$ and $$y$$ so try out $$cos~x + cos~y$$ for now. We've got the loop; now to check what that looks like when $$h = 0$$ so we can modify it for the square and woah we're already done.