Streamline, $\psi$ and $\phi$ for given field

Given: $u = 2x^3$, $v = -6x^2y$.

Check of continuity

$$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 6x^2 - 6x^2 = 0$$

so the flow is incompressible and a stream function exists.

Equation of streamline

$$\frac{dy}{dx} = \frac{v}{u} = \frac{-6x^2y}{2x^3} = -\frac{3y}{x}$$ $$\frac{dy}{y} = -3\frac{dx}{x}\;\Rightarrow\; \ln y = -3\ln x + \ln C\;\Rightarrow\; x^3y = C$$

Stream function ψ

$u = \partial\psi/\partial y = 2x^3\Rightarrow\psi = 2x^3y + f(x)$.

$v = -\partial\psi/\partial x = -6x^2y - f'(x)$. Comparing with $v = -6x^2y$ gives $f'(x) = 0$.

$$\psi = 2x^3y\ (+\text{constant})$$

This agrees with the streamline $x^3y = C$.

Velocity potential φ

A velocity potential exists only for irrotational flow, i.e. when $\partial v/\partial x = \partial u/\partial y$.

$$\frac{\partial v}{\partial x} = -12xy,\qquad \frac{\partial u}{\partial y} = 0$$

The vorticity $\omega_z = \tfrac12(\partial v/\partial x - \partial u/\partial y) = -6xy \ne 0$, so the flow is rotational.

Answer: streamline $x^3y = C$; $\psi = 2x^3y$; $\phi$ does not exist (flow is rotational, $\omega_z = -6xy$).