(a) Differentiate the equation \(x^3 + y^2 + 3x^2 + 3y = 4\) implicitly with respect to \(x\):

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

Rearrange to solve for \(\frac{dy}{dx}\):

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

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

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

(b) For the tangent to be parallel to the x-axis, \(\frac{dy}{dx} = 0\).

Set the numerator of \(\frac{dy}{dx} = -\frac{3x^2 + 6x}{2y + 3}\) to zero:

\(3x^2 + 6x = 0\).

Factorize:

\(3x(x + 2) = 0\).

\(x = 0\) or \(x = -2\).

Substitute \(x = 0\) into the original equation:

\(y^2 + 3y - 4 = 0\).

Factorize:

\((y - 1)(y + 4) = 0\).

\(y = 1\) or \(y = -4\).

Coordinates: \((0, 1)\) and \((0, -4)\).

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

\(y^2 + 3y = 0\).

Factorize:

\(y(y + 3) = 0\).

\(y = 0\) or \(y = -3\).

Coordinates: \((-2, 0)\) and \((-2, -3)\).

To find the gradient of the curve \(x^3 + 3xy^2 - y^3 = 1\), we need to differentiate implicitly with respect to \(x\).

The derivative of \(x^3\) is \(3x^2\).

For \(3xy^2\), use the product rule: \(3(y^2 + 2xy\frac{dy}{dx}) = 3y^2 + 6xy\frac{dy}{dx}\).

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

Substituting these into the equation, we get:

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

Simplify to find \(\frac{dy}{dx}\):

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

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

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

Substitute \(x = 1\) and \(y = 3\):

\(\frac{dy}{dx} = \frac{-3(1)^2 - 3(3)^2}{6(1)(3) - 3(3)^2}\).

\(\frac{dy}{dx} = \frac{-3 - 27}{18 - 27}\).

\(\frac{dy}{dx} = \frac{-30}{-9} = \frac{10}{3}\).

(i) Differentiate the equation \(2x^3 - y^3 - 3xy^2 = 2a^3\) implicitly with respect to \(x\):

\(\frac{d}{dx}(2x^3) = 6x^2\)

\(\frac{d}{dx}(-y^3) = -3y^2 \frac{dy}{dx}\)

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

Combine and simplify:

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

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

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

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

(ii) For the tangent to be parallel to the y-axis, \(\frac{dx}{dy} = 0\), which implies \(\frac{dy}{dx}\) is undefined. This occurs when the denominator \(y^2 + 2xy = 0\).

Solve \(y^2 + 2xy = 0\):

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

\(y = 0\) or \(y = -2x\)

For \(y = 0\), substitute into the curve equation:

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

\(x = a\)

Point: \((a, 0)\)

For \(y = -2x\), substitute into the curve equation:

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

\(2x^3 + 8x^3 - 12x^3 = 2a^3\)

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

\(x = -a\)

\(y = -2(-a) = 2a\)

Point: \((-a, 2a)\)

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