(a)(i) The gradient of the normal at \(P\) is \(-\frac{1}{\frac{dy}{dx}}\). The length \(MN\) is the horizontal distance from \(M\) to \(N\), which is \(y \cdot \frac{dy}{dx}\). Therefore, \(\frac{MN}{y} = \frac{dy}{dx}\).

(a)(ii) The area of triangle \(PMN\) is \(\frac{1}{2} \times MN \times y = \frac{1}{2} \times y \cdot \frac{dy}{dx} \times y = \frac{1}{2}y^2 \frac{dy}{dx}\). Given that this area equals \(\tan x\), we have \(\frac{1}{2}y^2 \frac{dy}{dx} = \tan x\).

(b) Separate variables: \(\frac{1}{2}y^2 dy = \tan x \, dx\).

Integrate both sides: \(\int \frac{1}{2}y^2 dy = \int \tan x \, dx\).

This gives \(\frac{1}{6}y^3 = -\ln |\cos x| + C\).

Using the initial condition \(y = 1\) when \(x = 0\), we find \(C = \frac{1}{6}\).

Thus, \(\frac{1}{6}y^3 = -\ln |\cos x| + \frac{1}{6}\).

Rearranging gives \(y^3 = 1 - 6\ln |\cos x|\).

Therefore, \(y = \sqrt[3]{1 - 6\ln \cos x}\).