To find the minimum point of the curve \(y = \frac{9}{4}x^2 - 12x + 18\), we can complete the square or use calculus.

First, we find the derivative \(\frac{dy}{dx} = \frac{9}{2}x - 12\).

Setting \(\frac{dy}{dx} = 0\) gives \(\frac{9}{2}x - 12 = 0\).

Solving for \(x\), we get \(x = \frac{24}{9} = \frac{8}{3}\).

Substitute \(x = \frac{8}{3}\) back into the original equation:

\(y = \frac{9}{4} \left( \frac{8}{3} \right)^2 - 12 \left( \frac{8}{3} \right) + 18\).

Calculate \(y = \frac{9}{4} \times \frac{64}{9} - 32 + 18\).

\(y = 16 - 32 + 18 = 2\).

Thus, the coordinates of the minimum point are \(\left( \frac{8}{3}, 2 \right)\).