I'm extremely confused about which way the gradient points. If given the equation $z = x^3+y^3-6xy$, I could calculate the gradient at point (1, 2, -3) by rearranging the equation to $f(x,y,z) = x^3+y^3-6xy -z = 0$, and get a gradient of (-9, 6, -1).
But i could also rearrange to $f(x,y,z) = z-x^3-y^3+6xy = 0$, and get a gradient of (9, -6, 1), which points in the opposite direction. How do I know which way the gradient is supposed to point? Is rearranging in one of these ways wrong?
The gradient of a function of 2 variables $f(x,y)$ is $(\partial f/\partial x, \partial f/\partial y)$. In your case you happen to have used $z$ as the symbol for your function so the gradient is $(\partial z/\partial x, \partial z/\partial y)$.
Now, it is another matter if you want to find a normal vector to some surface. The normal is parallel to the gradient vector of the level surfaces. It doesn't matter which way you re-arrange your function in to a level surface. Yes the 'gradients' (i.e. the vector of partial derivatives) point in opposite directions but this doesn't matter because they are parallel and either are suitable as a normal vector. In other words, if $\vec{n}$ is a normal vector, then so is $-\vec{n}$.