In two-dimensional, incompressible flow we can define a stream function $\psi$ by a line integral along any path $C$ joining $(0,0)$ to $(x,y)$
$$\psi(x,y) = \int_C \mathbb{u} \cdot \mathbb{n} dl = \int_C u \, dy - v \, dx.$$
Using the incompressiblity condition $\nabla \cdot \mathbb{u} = 0$ and Green's theorem it follows that the line integral is independent of path and $\psi$ is a well-defined function which is related to the velocity field by
$$u = \frac{\partial \psi}{\partial y}, \,\,\,\ v = -\frac{\partial \psi}{\partial x}. $$
Hence,
$$\mathbb{u} \cdot \nabla \psi = u \frac{\partial \psi}{\partial x} + v \frac{\partial \psi}{\partial y}= u(-v) + v(u) = 0.$$
The gradient $\nabla \psi$ is always oriented in a direction normal to a level curve $\psi = \text{constant}.$ The condition $\mathbb{u} \cdot \nabla \psi = 0$ implies that the velocity at a point is tangential to such a curve. Hence, level curves of $\psi$ correspond to streamlines.
We can relate the stream function and vorticity, using the definition $\mathbb{\omega} = \nabla \times \mathbb{u}$ by
$$\xi = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} = \frac{\partial }{\partial x}\left(-\frac{\partial \psi}{\partial x}\right) - \frac{\partial }{\partial y}\left(\frac{\partial \psi}{\partial y}\right) = -\nabla^2 \psi.$$
In steady inviscid flow we have the Euler equation
$$\mathbb{u} \cdot \nabla \mathbb{u} = -\frac{1}{\rho} \nabla p.$$
Taking the curl of both sides we find
$$\mathbb{u} \cdot \nabla \mathbb{\mathbb{\omega}} + \mathbb{\omega} \cdot \nabla \mathbb{\mathbb{u}} = 0.$$
Since $\mathbb{\omega}$ has only the non-zero $z-$ component $\xi$ this reduces to
$$\mathbb{u} \cdot \nabla \xi = 0,$$
and, by the same argument for the stream function, te vorticity must be constant along a streamline.
The stream function exists for a two-dimensional incompressible flow and is unique up to a constant that can be set arbitrarily to $0$. Since the velocity components depend on partial derivatives of the stream function and the volumetric flow rate is the difference in stream function values at two points, the constant is irrelevant.
As you have done, starting with
$$- \frac{\partial \psi}{\partial r} = u_\theta = -U \sin \theta,$$
and integrating, we get
$$\psi = Ur \sin \theta + g(\theta)$$
Upon differentiating with respect to $\theta$ we find
$$U \cos\theta = \frac{1}{r}\frac{\partial\psi}{\partial \theta} = \frac{1}{r}\frac{\partial}{\partial \theta}\left(Ur \sin \theta + g(\theta) \right) = U \cos \theta + \frac{g'(\theta)}{r}$$
Thus, $g'(\theta) = 0$, which implies that $g(\theta) = C$ -- a constant that can be arbitrarily set to $0$ as discussed above. In this way we obtain $\psi(r,\theta) = U r \sin \theta$.
Best Answer
The given velocity field does not correspond to incompressible flow since the continuity equation is not satisfied, i.e.,
$$\nabla \cdot \mathbf{u} = \frac{\partial u }{\partial x} + \frac{\partial v}{\partial y} = \frac{\partial}{\partial x} (x) + \frac{\partial }{\partial y} (2y) = 3 \neq 0$$
The streamfunction does not exist in this case.
A proof of the existence of a streamfunction for two-dimensional, incompressible flow satisfying $\psi_y = u, \, \psi_x = -v$ is given here.
Aside
This velocity field is irrotational with
$$\nabla \times \mathbf{u} = \frac{\partial v }{\partial x} - \frac{\partial u}{\partial y} = 0,$$
and can be expressed in terms of a poatential
$$\mathbf{u} = - \nabla \phi,$$
where $\phi = -x^2/2 - y^2 +C$.