(a) To find the stationary point, we need to find \(\frac{dy}{dx}\) and set it to zero. Using the quotient rule, \(\frac{dy}{dx} = \frac{(9-x)(-\frac{1}{2}x^{-3/2}) - \sqrt{x}}{(9-x)^2 x}\).

Simplifying, \(\frac{dy}{dx} = \frac{-\frac{1}{2}(9-x)x^{-3/2} - x^{-1/2}}{(9-x)^2}\).

Setting \(\frac{dy}{dx} = 0\), we solve for \(x\) to get \(x = 3\).

(b) Using the substitution \(u = \sqrt{x}\), then \(x = u^2\) and \(dx = 2u \, du\).

The integral becomes \(\int_0^2 \frac{1}{(9-u^2)u} \, 2u \, du = \int_0^2 \frac{2}{9-u^2} \, du\).

This can be integrated using partial fractions or a standard integral formula to obtain \(\frac{1}{3} \ln \left( \frac{3+u}{3-u} \right)\).

Evaluating from \(u = 0\) to \(u = 2\), we get \(\frac{1}{3} \ln 5\).