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

(i) Differentiate the equation \(2x^4 + xy^3 + y^4 = 10\) implicitly with respect to \(x\):

\(\frac{d}{dx}(2x^4) = 8x^3\).

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

For \(y^4\), use the chain rule: \(\frac{d}{dx}(y^4) = 4y^3 \frac{dy}{dx}\).

Substitute these into the differentiated equation:

\(8x^3 + y^3 + 3xy^2 \frac{dy}{dx} + 4y^3 \frac{dy}{dx} = 0\).

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

\(3xy^2 \frac{dy}{dx} + 4y^3 \frac{dy}{dx} = -8x^3 - y^3\).

\(\frac{dy}{dx} = -\frac{8x^3 + y^3}{3xy^2 + 4y^3}\).

(ii) For the tangent to be parallel to the x-axis, \(\frac{dy}{dx} = 0\). Set the numerator of \(\frac{dy}{dx}\) to zero:

\(-8x^3 - y^3 = 0\) or \(y^3 = -8x^3\).

\(y = -2x\).

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

\(2x^4 + x(-2x)^3 + (-2x)^4 = 10\).

\(2x^4 - 8x^4 + 16x^4 = 10\).

\(10x^4 = 10\).

\(x^4 = 1\).

\(x = 1\) or \(x = -1\).

For \(x = 1\), \(y = -2\).

For \(x = -1\), \(y = 2\).

The points are \((-1, 2)\) and \((1, -2)\).

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

Start by differentiating the given equation \(xy(x - 6y) = 9a^3\).

Using the product rule, differentiate the left-hand side:

\(\frac{d}{dx}[xy(x - 6y)] = x^2 \frac{dy}{dx} + 2xy \frac{dy}{dx} + y(x - 6y)\).

Equate the derivative to zero for the tangent to be horizontal:

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

Simplify and solve for \(\frac{dy}{dx} = 0\):

\(y(x - 6y) = 0\).

This gives \(y = 0\) or \(x - 6y = 0\).

Since \(y = 0\) does not satisfy the original equation, we use \(x - 6y = 0\).

Substitute \(x = 6y\) into the original equation:

\((6y)y(6y - 6y) = 9a^3\).

\(6y^2 = 9a^3\).

\(y = -a\).

Substitute \(y = -a\) back into \(x = 6y\):

\(x = 6(-a) = -3a\).

Thus, the coordinates are \((-3a, -a)\).

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

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

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

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

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

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

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

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

\(x^2 - 2xy = 0\).

Factor to find \(x(x - 2y) = 0\), giving \(x = 0\) or \(x = 2y\).

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

\(y^3 = 3\), so \(y = 1.44\).

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

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

\(8y^3 - 12y^3 + y^3 = 3\).

\(-3y^3 = 3\), so \(y = -1\) and \(x = -2\).

The points are \((-2, -1)\) and \((0, 1.44)\).

(i) Differentiate both sides of \(\sin y \ln x = x - 2 \sin y\) with respect to \(x\).

Using the product rule on the left side, we have:

\(\frac{d}{dx}(\sin y \ln x) = \cos y \frac{dy}{dx} \ln x + \sin y \frac{1}{x}\).

The derivative of the right side is:

\(1 - 2\cos y \frac{dy}{dx}\).

Equating the derivatives, we get:

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

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

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

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

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

Set the numerator of \(\frac{dy}{dx}\) to zero:

\(1 - \frac{\sin y}{x} = 0\).

\(x = \sin y\).

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

\(\sin y \ln \sin y = \sin y - 2\sin y\).

\(\ln \sin y = -1\).

\(\sin y = \frac{1}{e}\).

Thus, \(x = \frac{1}{e}\).