Given the volume of water \(V = \pi \left( \frac{1}{2}h^2 + h \right)\), differentiate with respect to \(h\) to find \(\frac{dV}{dh}\).

\(\frac{dV}{dh} = \pi (h + 1)\).

We know \(\frac{dV}{dt} = 2\) cm\(^3\) s\(^{-1}\). We need to find \(\frac{dh}{dt}\) when \(h = 3\) cm.

Using the chain rule: \(\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}\).

Substitute \(\frac{dV}{dh} = \pi (h + 1)\) and \(\frac{dV}{dt} = 2\):

\(2 = \pi (h + 1) \cdot \frac{dh}{dt}\).

When \(h = 3\), \(\pi (3 + 1) = 4\pi\).

\(2 = 4\pi \cdot \frac{dh}{dt}\).

\(\frac{dh}{dt} = \frac{2}{4\pi} = \frac{1}{2\pi}\) cm s\(^{-1}\).