(i) Differentiate \(\ln(xy)\) implicitly: \(\frac{1}{xy}(y + x\frac{dy}{dx}) = \frac{1}{x} + \frac{1}{y}\frac{dy}{dx}\).

Differentiate \(y^3\): \(3y^2\frac{dy}{dx}\).

Set the derivatives equal: \(\frac{1}{x} + \frac{1}{y}\frac{dy}{dx} = 3y^2\frac{dy}{dx}\).

Solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{y}{x(3y^3 - 1)}\).

(ii) The tangent is parallel to the y-axis when \(\frac{dy}{dx}\) is undefined, i.e., when \(3y^3 - 1 = 0\).

Solve \(3y^3 - 1 = 0\): \(y = 0.693\).

Substitute \(y = 0.693\) into the original equation: \(\ln(x \times 0.693) - (0.693)^3 = 1\).

Solve for \(x\): \(x = 5.47\).

(i) To find the gradient, we need to differentiate the equation implicitly with respect to \(x\):

\(\frac{d}{dx}(3x^2) - \frac{d}{dx}(4xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(45)\).

This gives \(6x - (4y + 4x\frac{dy}{dx}) + 2y\frac{dy}{dx} = 0\).

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

\(6x - 4y = (4x - 2y)\frac{dy}{dx}\).

\(\frac{dy}{dx} = \frac{6x - 4y}{4x - 2y}\).

Substitute \(x = 2\) and \(y = -3\):

\(\frac{dy}{dx} = \frac{6(2) - 4(-3)}{4(2) - 2(-3)} = \frac{12 + 12}{8 + 6} = \frac{24}{14} = \frac{12}{7}\).

(ii) To show there are no points where the gradient is 1, set \(\frac{dy}{dx} = 1\):

\(\frac{6x - 4y}{4x - 2y} = 1\).

This implies \(6x - 4y = 4x - 2y\).

Simplifying gives \(2x = 2y\) or \(y = x\).

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

\(3x^2 - 4x^2 + x^2 = 45\).

This simplifies to \(0 = 45\), a contradiction.

Therefore, there are no points on the curve where the gradient is 1.

(i) Start with the equation \(x \ln y = 2x + 1\).

Differentiate both sides with respect to \(x\):

\(\frac{d}{dx}(x \ln y) = \frac{d}{dx}(2x + 1)\).

Using the product rule on the left side: \(\ln y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx} = 2\).

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

\(x \cdot \frac{1}{y} \cdot \frac{dy}{dx} = 2 - \ln y\).

\(\frac{dy}{dx} = \frac{y(2 - \ln y)}{x}\).

Substitute \(\ln y = \frac{2x + 1}{x}\) from the original equation:

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

(ii) Given \(y = 1\), substitute into the original equation:

\(x \ln 1 = 2x + 1 \Rightarrow 0 = 2x + 1 \Rightarrow x = -\frac{1}{2}\).

Find the gradient at this point using \(\frac{dy}{dx} = -\frac{y}{x^2}\):

\(\frac{dy}{dx} = -\frac{1}{(-\frac{1}{2})^2} = -4\).

The equation of the tangent line is \(y - 1 = -4(x + \frac{1}{2})\).

Simplify to the form \(ax + by + c = 0\):

\(y - 1 = -4x - 2 \Rightarrow y + 4x + 1 = 0\).

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

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

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

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

Set the derivative equal to zero:

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

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

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

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

(ii) To find the equation of the tangent at \((2, 1)\), first find the gradient:

Substitute \(x = 2\) and \(y = 1\) into \(\frac{dy}{dx} = \frac{3x^2 - 2xy}{x^2 + 3y^2}\):

\(\frac{dy}{dx} = \frac{3(2)^2 - 2(2)(1)}{(2)^2 + 3(1)^2} = \frac{12 - 4}{4 + 3} = \frac{8}{7}\)

The equation of the tangent line is:

\(y - 1 = \frac{8}{7}(x - 2)\)

Rearrange to the form \(ax + by + c = 0\):

\(7(y - 1) = 8(x - 2)\)

\(7y - 7 = 8x - 16\)

\(8x - 7y - 9 = 0\)

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 \((a, -2a)\).