Finding the trajectory given the direction of travel and initial position 
Question Given $f(x,y) = x^{4} - 6x^{2}y^{2} + y^{4} - 2x^{2} + 2y^{2}$.
Suppose a particle moves along the surface $z = f(x,y)$ such that it always travels in the direction in which $f$ decreases most rapidly. Given the particle starts at the point $(x,y) = (2,1)$, show that it will follow a path on which
$$ xy(x^{2} - y^{2} - 1) = \text{constant}$$

Computing the directional derivative of this function, I believe the particle moves in the direction of the vector $\vec{u}$ which is given by
$$ \vec{u} = -\frac{\nabla f}{| \nabla f |} = -\frac{(4x^{3} - 12xy^{2} - 4x, -12x^{2}y + 4y^{3} + 4y )}{| \nabla f |} $$
So that, initially $\vec{u} = (0,1)$ and the particle moves along the y-axis.
How can I then use this information to determine the surface on which the particle moves?
 A: Here is a Mathematica visualization that may be appreciated. First, we make a shaded contour plot of the given $f(x,y)$; I then additionally include a few blue contours of the indicated curves of steepest descent, including one passing through the starting point $(2,1)$ marked in red.
Show[
  ContourPlot[x^4 - 6 x^2 y^2 + y^4 - 2 x^2 + 2 x^2, {x, 0, 2.2}, {y, 0, 1.4},
     Contours -> 10, ContourShading -> False, ContourStyle -> Red], 
  ContourPlot[x y (x^2 - y^2 - 1), {x, 0, 2.2}, {y, 0, 1.5}, 
     ContourShading -> False, ContourStyle -> Directive[Blue, Thick], 
     Contours -> {0, 1, 2, 3, 4}], 
  Graphics[{Red, PointSize[.021], Point[{2, 1}]}]]


A: To get the right direction, you don't need to normalize. The gradient gives you the direction of the steepest slope, so from:
$$\nabla f = \left( \,f_x \,,\, f_y \right) = \left( 4x^{3} - 12xy^{2} - 4x, -12x^{2}y + 4y^{3} + 4y \right)$$
you want:
$$\frac{\mbox{d}y}{\mbox{d}x} = \frac{f_y}{f_x} = \frac{-12x^{2}y + 4y^{3} + 4y}{4x^{3} - 12xy^{2} - 4x} = \frac{-3x^{2}y + y^{3} + y}{x^{3} - 3xy^{2} - x} \tag{$*$}$$
Now you can use implicit differentiation to show that the trajectory given by:

$$ xy(x^{2} - y^{2} - 1) = \text{constant}$$

satisfies $(*)$.
