Two functions are orthogonal trajectories of each other if their gradients multiply together to give -1.

The gradient of `y=ax^3` is `(dy)/(dx) = 3ax^2`

Using implicit differentiation, `d/(dx) (x^2 +3y^2 - b) = 0`

`implies 2x + 6y*(dy)/(dx) - 0 = 0`

`implies 6y*(dy)/(dx) = -2x`

`implies (dy)/(dx) = -x/(3y)`

Multiplying the two gradients together we get `3ax^2(-x/(3y)) = -(ax^3)/y`

Now, from the equation `y = ax^3` we can substitute in for `y` giving a product of gradients of `-(ax^3)/(ax^3) = -1` **QED**