A glass of water is rotating about its axis at constant angular velocity $\omega$. Let $y=f(x)$ denote the equation of the curve obtained by cutting the surface of the liquid with a plane passing through its axis of rotation.
Show that $f'(x) = \frac{\omega^2}{g}x$, where $g$ is the acceleration of free fall.
I don't understand where the plane is or what $f$ represents.
I will assume that the glass is upright with its axis of rotation passing through the origin. Is $f$ the height of the water at a cross section that is through the centre of the glass?
The glass is rotating about that vertical line, and so the water bulges as such. The plane passes through as in this image. $f$ is the equation of that bulge, so yes, the height of the water at the cross section.