Related to https://math.stackexchange.com/questions/1667643/steepest-part-of-cos-x-cos-y
Is there a general way of finding the steepest part of a surface? I know that to find the steepest part of a normal function $f$, you'd look for solutions $f'' = 0$, but for surfaces you're using the gradient, not differentiation, but $\nabla^2\mathbf f = \mathbf0$ doesn't seem to be the correct way to proceed. Is there another way?
I know about the steepest direction of a point on a surface is $\mathbf{u} = \dfrac{\nabla\mathbf f(a)}{|\nabla\mathbf f(a)|}$, but I don't think this is very helpful in this case?
The norm of the gradient of a function gives the steepness you are looking for, so we are really trying to maximize this norm. Of course, the norm is not differentiable at zero, so better maximize the norm squared. Thus, you should solve $\nabla \Vert\nabla f (x,y)\Vert^2=0.$