The area \( A \) of an equilateral triangle with side \( s \) is given by the formula:
\[
A = \frac{\sqrt{3}}{4} s^2
\]
We are given that \( \frac{ds}{dt} = 3 \, \text{cm/s} \) and \( s = 15 \, \text{cm} \), and we need to find \( \frac{dA}{dt} \).
Differentiate both sides of the area formula with respect to time \( t \):
\[
\frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2s \cdot \frac{ds}{dt}
\]
Substitute \( s = 15 \) and \( \frac{ds}{dt} = 3 \):
\[
\frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2 \cdot 15 \cdot 3 = \frac{\sqrt{3}}{4} \cdot 90 = 75 \, \text{cm}^2/\text{s}
\]
Thus, the rate of change of the area is 75 cm\(^2\)/s.