(i) To find the maximum point \(M\), we need to find the derivative of the function \(y = e^{-x} - e^{-2x}\) and set it to zero.

The derivative is \(\frac{dy}{dx} = -e^{-x} - (-2)e^{-2x} = -e^{-x} + 2e^{-2x}\).

Setting the derivative to zero gives:

\(-e^{-x} + 2e^{-2x} = 0\)

\(e^{-x} = 2e^{-2x}\)

\(e^{x} = 2\)

\(x = \ln 2\)

Thus, \(p = \ln 2\).

(ii) To find the area of the shaded region, we need to integrate the function from \(x = 0\) to \(x = p\).

The indefinite integral of \(y = e^{-x} - e^{-2x}\) is:

\(\int (e^{-x} - e^{-2x}) \, dx = -e^{-x} - \left(-\frac{1}{2}\right)e^{-2x} = -e^{-x} + \frac{1}{2}e^{-2x}\)

Evaluating from \(x = 0\) to \(x = \ln 2\):

\(\left[-e^{-x} + \frac{1}{2}e^{-2x}\right]_0^{\ln 2} = \left(-e^{-\ln 2} + \frac{1}{2}e^{-2\ln 2}\right) - \left(-e^{0} + \frac{1}{2}e^{0}\right)\)

\(= \left(-\frac{1}{2} + \frac{1}{2}\left(\frac{1}{4}\right)\right) - \left(-1 + \frac{1}{2}\right)\)

\(= \left(-\frac{1}{2} + \frac{1}{8}\right) - \left(-\frac{1}{2}\right)\)

\(= -\frac{1}{2} + \frac{1}{8} + \frac{1}{2}\)

\(= \frac{1}{8}\)

Thus, the area is \(\frac{1}{8}\).