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

Start by differentiating the given equation \(xy(x+y) = 2a^3\).

Using the product rule, differentiate \(xy(x+y)\):

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

Set \(\frac{dy}{dx} = 0\) for the tangent to be parallel to the \(x\)-axis:

\(y^2 + 2xy = 0\).

Factor the equation:

\(y(y + 2x) = 0\).

Since \(y = 0\) is not possible (as it would make the original equation undefined), we have:

\(y = -2x\).

Substitute \(y = -2x\) into the original equation:

\(x(-2x)(x - 2x) = 2a^3\).

\(-2x^3 = 2a^3\).

Solve for \(x\):

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

\(x = a\).

Substitute \(x = a\) back into \(y = -2x\):

\(y = -2a\).

Thus, the coordinates of the point are \((a, -2a)\).