Start by marking the stationary points.
Next, get the gradient vector field
∇z = 〈 ∂z/∂x , ∂z/∂y 〉 = ∇z = 〈 3x2 - 12x , -2y 〉
The gradient is the zero vector at the two stationary points.
At any other point P, the gradient vector at P is perpendicular to the contour line that goes through P.
Get the differential equation for which the solution set is the set of all contours. That is, set z(x,y) = Constant, and find dy/dx by implicit differentiation.
x3 - 6x2 - y2 = C
3x2 - 12x - 2y dy/dx = 0
dy/dx = (3x2 - 12x) / (2y) = 3x(x - 4) / (2y)
Now evaluate dy/dx at many different points to draw a slope field.
At any point P, the contour through P must have the slope equal to the slope assigned at P.
Tip: knowing the points where dy/dx = 0 (horizontal slope), and dy/dx = undefined (vertical slope) is useful here. The slope marks at such points are called nullclines.