The way to proceed is as follows. Take a coordinate system with $x$ pointing to the right and $y$ pointing down. Imagine the inverted bell-shaped drop hanging from the $y = 0$ plane. Now "cut" the bell horizontally at at $y$. The circumference of the bell at the cut is $2\pi x$ and the total force on the cut part is $2\pi xc$, where $c$ is the surface tension per unit length (a constant). The surface tension is parallel to the tangent and the total upwards force on the cut part has the component $$F_y = -2\pi xc\sin(a),$$ where $a$ is the angle (a negative value) of the tangent at $(x,y)$.
Now, this force must equal the weight of the cut part of the bell. The volume $V$ of this part is the integral
$$V = \int_y^{y_{tip}}\pi x^2dy.$$
The mass of this volume is $mV$, where $m$ is the specific mass, and the force of gravity thus $gmV$. Hence the basic equation is $$-2pcx\sin(a) = gmV.$$
Now, $V$ is an integral of $x^2$ versus $y$ and $\tan(a) = dy/dx = y'$, so $$sin(a) = \frac{y'}{\sqrt{1+y'^2}}.$$
Differentiating with respect to $y$ on both sides gives the final differential equation:
$$2\pi c\frac{d(xy'/\sqrt{1+y'^2})}{dy} = gm\pi x^2$$
or
$$\frac{d(xy'/\sqrt{1+y'^2})}{dy} = kx^2,$$
where $k = gm/2c$. This can be solved exactly by a change of variables, but I wont go into the technicalities (I don't have a math texter) but just say that the shape of the bell (surface of the drop) involves elliptic functions.
Further, if you add the acceleration $mV\,d^2Y/dt^2$ (where $Y$ is the center of gravity of the drop, which is simple to express from the quasi-stationary shape) to the force difference at $y = 0$ and slowly increase its volume at a constant rate $dV/dt$, then you get an equation which can be solved to give the time of pinch-off, where the drop falls, as well as the maximum volume the surface tension can support. One day I will repeat the math, which I actually did in my student days.
