(a) Substitute \(x = r \cos \theta\) and \(y = r \sin \theta\) into the Cartesian equation \((x^2 + y^2)^2 = 6xy\).

This gives \((r^2)^2 = 6r^2 \cos \theta \sin \theta\).

Thus, \(r^4 = 6r^2 \cos \theta \sin \theta\).

Divide both sides by \(r^2\) (assuming \(r \neq 0\)) to get \(r^2 = 6 \cos \theta \sin \theta\).

Using the identity \(\sin 2\theta = 2 \sin \theta \cos \theta\), we have \(r^2 = 3 \sin 2\theta\).

(b) The sketch shows a single loop symmetrical about \(\theta = \frac{1}{4}\pi\). The maximum distance from the pole is \(\sqrt{3}\).

(c) The area enclosed by \(C\) is given by \(\frac{1}{2} \int_0^{\frac{1}{2}\pi} r^2 \, d\theta\).

Substitute \(r^2 = 3 \sin 2\theta\) to get \(\frac{3}{2} \int_0^{\frac{1}{2}\pi} \sin 2\theta \, d\theta\).

Integrate to find \(\frac{3}{2} \left[ -\frac{1}{2} \cos 2\theta \right]_0^{\frac{1}{2}\pi} = \frac{3}{2}\).

(d) To find the maximum distance from the initial line, set \(y = r \sin \theta = 3^{\frac{3}{4}} \sin^{\frac{1}{2}} 2\theta \sin \theta\).

Set \(\frac{dy}{d\theta} = 0\) and solve \(\sin^2 2\theta \cos \theta + \sin^{-\frac{1}{2}} 2\theta \cos 2\theta \sin \theta = 0\).

Using trigonometric identities, solve for \(\theta = \frac{1}{3}\pi\).

The maximum distance is \(\frac{3^{\frac{3}{4}}}{2^{\frac{1}{2}}} = 1.40\).