To find where the tangent is parallel to the \(x\)-axis, we need \(\frac{dy}{dx} = 0\).

Start by differentiating the given equation \(xy(x - 6y) = 9a^3\).

Using the product rule, differentiate the left-hand side:

\(\frac{d}{dx}[xy(x - 6y)] = x^2 \frac{dy}{dx} + 2xy \frac{dy}{dx} + y(x - 6y)\).

Equate the derivative to zero for the tangent to be horizontal:

\(x^2 \frac{dy}{dx} + 2xy \frac{dy}{dx} + y(x - 6y) = 0\).

Simplify and solve for \(\frac{dy}{dx} = 0\):

\(y(x - 6y) = 0\).

This gives \(y = 0\) or \(x - 6y = 0\).

Since \(y = 0\) does not satisfy the original equation, we use \(x - 6y = 0\).

Substitute \(x = 6y\) into the original equation:

\((6y)y(6y - 6y) = 9a^3\).

\(6y^2 = 9a^3\).

\(y = -a\).

Substitute \(y = -a\) back into \(x = 6y\):

\(x = 6(-a) = -3a\).

Thus, the coordinates are \((-3a, -a)\).