Wikipedia references:
Streamlines are a family of curves that are instantaneously tangent to the velocity vector of the flow. These show the direction a massless fluid element will travel in at any point in time.
Consider the velocity field
$(u,v)$ of a two-dimensional incompressible flow. Let the family of curves be given by
$\;\psi(x,y) = c$ . The velocity vectors are tangent to these as shown for one of them in the following picture.

Thus, along the curve $\psi(x,y) = c$ , the following equations hold: $ \left. \begin{array}{l} \frac{dy}{dx} = \frac{v}{u} \\ d\psi = 0 = \frac{\partial \psi}{\partial x} dx + \frac{\partial \psi}{\partial y} dy \end{array} \right\} \qquad \Longrightarrow \qquad \frac{dy}{dx} = - \frac{\partial \psi / \partial x}{\partial \psi / \partial y} = \frac{v}{u} $ Hence, apart from a constant: $ u = \frac{\partial \psi}{\partial y} \qquad ; \qquad v = - \frac{\partial \psi}{\partial x} $ But the flow is incompressible, so: $ \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0 \qquad \Longrightarrow \qquad \frac{\partial^2 \psi}{\partial x\, \partial y} = \frac{\partial^2 \psi}{\partial y\, \partial x} $ Herewith the conditions for an exact differential equation are fulfilled. Now solve $\psi$ from: $ v\, dx - u\, dy = 0 $ Example. As taken from : Find the velocity of a flow . $ u = -\frac{y}{x^2+y^2} \qquad ; \qquad v = \frac{x}{x^2+y^2} $ Then: $ v\, dx - u\, dy = \frac{x\,dx + y\,dy}{x^2+y^2} = \frac{d\left( x^2+y^2 \right)}{x^2+y^2} = 0 \qquad \Longrightarrow \qquad x^2 + y^2 = c $ It is concluded that the streamlines of this flow are circles.
Example. Somewhat related to the above one. $ u = \lambda\,x \qquad ; \qquad v = \lambda\,y $ Then, assuming that $\; x\ne 0$ (i.e. $\,x=0\,$ as a special case) : $ v\, dx - u\, dy = 0 \quad \Longleftrightarrow \quad \frac{y\,dx - x\,dy}{x^2} = - d(y/x) = 0 \quad \Longrightarrow \quad y = c\, x $ An integrating factor has been used. It is concluded that the streamlines of this flow are straight lines through the origin.
Wikipedia reference:
The electric potential at a point
$\vec{r}$ in a two-dimensional static electric field
$\vec{E}$ is given by the line integral:
$ V = - \int_C \vec{E}\cdot d\vec{r} = - \int_C \left(E_x\, dx + E_y\, dy\right) $ where
$C$ is an arbitrary path connecting the point with zero potential to
$\vec{r}$. It follows that:
$ E_x = - \frac{\partial V}{\partial x} \qquad ; \qquad E_y = - \frac{\partial V}{\partial y} $ The integral is zero if the path is closed. Then Green's theorem tells us:
$ \oint \left( E_x\, dx + E_y\, dy \right) = \iint \left( \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} \right) dx\,dy = - \iint \left( \frac{\partial^2 V}{\partial x \, \partial y} - \frac{\partial^2 V}{\partial y \, \partial x}\right) dx\,dy = 0 $ Thus establishing once more the
conditions for solvability of the exact differential equation:
$ E_x \, dx + E_y \, dy = 0 $ Solving this ODE results in the iso-lines
$\,V(x,y) = c\,$ of the electric potential
$\,V$ .
Example. An infinitely long and infinitely thin charged wire perpendicular to the plane and intersecting it in the origin. Apart from constants: $ (E_x,E_y) = \frac{(x,y)}{r^2} \quad \Longrightarrow \quad E_x\, dx + E_y\, dy = \frac{x\,dx + y\,dy}{r^2} = 0 \quad \Longrightarrow \quad x^2+y^2 = c $ The equipotential lines are circles.
Can of worms: special cases, and a singularity at the origin in all of the examples.