(i) Differentiate the equation \(x^2(x + 3y) - y^3 = 3\) implicitly with respect to \(x\).

The derivative of \(y^3\) is \(3y^2 \frac{dy}{dx}\).

The derivative of \(x^2(x + 3y)\) is \(2x(x + 3y) + x^2(1 + 3\frac{dy}{dx})\).

Equating the derivatives:

\(2x(x + 3y) + x^2(1 + 3\frac{dy}{dx}) - 3y^2 \frac{dy}{dx} = 0\)

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

\(2x^2 + 6xy + x^2 \frac{dy}{dx} + 3x^2 \frac{dy}{dx} - 3y^2 \frac{dy}{dx} = 0\)

\(2x^2 + 6xy = (3x^2 - 3y^2) \frac{dy}{dx}\)

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

(ii) The gradient of the normal is 1, so the gradient of the tangent is -1.

Set \(\frac{dy}{dx} = -1\):

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

\(x^2 + 2xy = -(y^2 - x^2)\)

\(x^2 + 2xy = -y^2 + x^2\)

\(2xy = -y^2\)

\(y = -2x\)

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

\(x^2(x + 3(-2x)) - (-2x)^3 = 3\)

\(x^2(x - 6x) + 8x^3 = 3\)

\(-5x^3 + 8x^3 = 3\)

\(3x^3 = 3\)

\(x^3 = 1\)

\(x = 1\)

\(y = -2\)

First point: \((1, -2)\)

For the second point, solve \(x^2 + 2xy = y^2 - x^2\) with \(y = 0\):

\(x^2 = 3\)

\(x = \sqrt[3]{3}\)

Second point: \((\sqrt[3]{3}, 0)\)