The orthogonal trajectories of u =$c_{1}$ are given by v = $c_{2}$ where v is the harmonic conjugate of u.

Given $u = 3x^2 y + 2x^2 - y^3 - 2y^2$

$u_{x} = \frac{\partial u}{\partial x} = 6xy + 4x \\ u_{y} = \frac{\partial u}{\partial y} = 3x^2 - 3y^2 - 4y$

Now $f'(z) = ux + ivx = ux - ivy$ (by Cauchy Riemann equation)

$f'(z) = 6xy + 4x - i(3x^2 - 3y^2 - 4y)$

By Milne Thompson method, we replace x to z, y to 0 (i.e. x = z, y = 0)

$f'(z) = 4z - i (3 - z^2)$

By integration

$f(z) = \frac{4z^2}{2} - i3\frac{z^3}{3} + c \\ f(z) = 2z^2 - iz^3 + c$

Now,

put z = x + iy

$u + iv = 2(x + iy)^2 - i(x + iy)^3 + c \\ u + iv = 2(x^2 + 2xyi - y^2) - i(x^3 + 3x^2 yi - 3xy^2 - iy^3) \\ = 2x^2 + 4xyi - 2y^2 - ix^3 + 3x^2 y + 3xy^2 i - y^3 + c \\ u + iv = (2x^2 - 2y^2 + 3x^2 y - y^3) + i(4xy - x^3 + 3xy^2) + c$

Imaginary part $v = 4xy - x^3 + 3xy^2$

Hence, the required orthogonal trajectories are $4xy - x^3 + 3xy^2$