(i) Differentiate the equation \(x^3 y - 3xy^3 = 2a^4\) implicitly with respect to \(x\):
\(\frac{d}{dx}(x^3 y) = 3x^2 y + x^3 \frac{dy}{dx}\)
\(\frac{d}{dx}(-3xy^3) = -3(y^3 + 3xy^2 \frac{dy}{dx})\)
Combine and simplify:
\(3x^2 y + x^3 \frac{dy}{dx} - 9xy^2 \frac{dy}{dx} - 3y^3 = 0\)
Rearrange to solve for \(\frac{dy}{dx}\):
\(x^3 \frac{dy}{dx} - 9xy^2 \frac{dy}{dx} = 3y^3 - 3x^2 y\)
\(\frac{dy}{dx}(x^3 - 9xy^2) = 3y^3 - 3x^2 y\)
\(\frac{dy}{dx} = \frac{3x^2 y - 3y^3}{9xy^2 - x^3}\)
(ii) For the tangent to be parallel to the \(x\)-axis, \(\frac{dy}{dx} = 0\).
Set the numerator of \(\frac{dy}{dx}\) to zero:
\(3x^2 y - 3y^3 = 0\)
\(3y(x^2 - y^2) = 0\)
\(y = 0\) or \(x^2 = y^2\)
For \(y = 0\), substitute into the original equation:
\(x^3(0) - 3x(0)^3 = 2a^4\), which is not possible.
For \(x^2 = y^2\), \(x = y\) or \(x = -y\).
Substitute \(x = y\) into the original equation:
\(x^3 x - 3x x^3 = 2a^4\)
\(x^4 - 3x^4 = 2a^4\)
\(-2x^4 = 2a^4\)
\(x^4 = -a^4\), which is not possible.
Substitute \(x = -y\):
\(x^3(-x) - 3x(-x)^3 = 2a^4\)
\(-x^4 + 3x^4 = 2a^4\)
\(2x^4 = 2a^4\)
\(x^4 = a^4\)
\(x = a\) or \(x = -a\)
Thus, the points are \((a, -a)\) and \((-a, a)\).
