(a) To find the equation of the curve, integrate \(\frac{dy}{dx} = 3x^{\frac{1}{2}} - 3x^{-\frac{1}{2}}\).

Integrating, we get:

\(y = \int (3x^{\frac{1}{2}} - 3x^{-\frac{1}{2}}) \, dx\)

\(y = \left[ \frac{3}{\frac{3}{2}} x^{\frac{3}{2}} \right] - \left[ \frac{3}{\frac{1}{2}} x^{\frac{1}{2}} \right] + c\)

\(y = 2x^{\frac{3}{2}} - 6x^{\frac{1}{2}} + c\)

Using the point \((3, 5)\) to find \(c\):

\(5 = 2(3)^{\frac{3}{2}} - 6(3)^{\frac{1}{2}} + c\)

\(5 = 2 \times 3\sqrt{3} - 6 \times \sqrt{3} + c\)

\(5 = 6\sqrt{3} - 6\sqrt{3} + c\)

\(c = 5\)

Thus, the equation is \(y = 2x^{\frac{3}{2}} - 6x^{\frac{1}{2}} + 5\).

(b) To find the \(x\)-coordinate of the stationary point, set \(\frac{dy}{dx} = 0\):

\(3x^{\frac{1}{2}} - 3x^{-\frac{1}{2}} = 0\)

\(3x^{\frac{1}{2}} = 3x^{-\frac{1}{2}}\)

\(x^{\frac{1}{2}} = x^{-\frac{1}{2}}\)

\(x = 1\)

(c) \(y\) increases as \(x\) increases when \(\frac{dy}{dx} > 0\):

\(3x^{\frac{1}{2}} - 3x^{-\frac{1}{2}} > 0\)

\(x^{\frac{1}{2}} > x^{-\frac{1}{2}}\)

\(x > 1\)