(a) To find \(\frac{dy}{dx}\), differentiate \(y = \frac{2}{(3 - 2x)^2} - x\). Using the chain rule, \(\frac{d}{dx} \left( \frac{2}{(3 - 2x)^2} \right) = 8x(3-2x)^{-3}\). The derivative of \(-x\) is \(-1\). Thus, \(\frac{dy}{dx} = \frac{8}{(3-2x)^3} - 1\).

For \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx}\) again: \(\frac{d^2y}{dx^2} = -3 \times 8 \times (3-2x)^{-4} \times (-2) = \frac{48}{(3-2x)^4}\).

To integrate \(y\), \(\int y \, dx = \int \left( \frac{2}{(3 - 2x)^2} - x \right) \, dx = \frac{1}{3-2x} - \frac{x^2}{2} + c\).

(b) To find the \(x\)-coordinate of \(M\), set \(\frac{dy}{dx} = 0\): \(0 = \frac{8}{(3-2x)^3} - 1\). Solving gives \(x = \frac{1}{2}\).

(c) The area under the curve is \(\int_{0}^{1/2} \left( \frac{2}{(3 - 2x)^2} - x \right) \, dx\). Using the integral from part (a) and evaluating from 0 to \(\frac{1}{2}\), the area is \(\frac{1}{24}\).