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

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

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

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

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

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

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

Set \(y - x^2 = 0\) which gives \(y = x^2\).

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

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

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

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

\(x^3(x^3 - 1) = 0\).

\(x^3 = 0\) or \(x^3 = 1\).

\(x = 0\) or \(x = 1\).

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

Thus, the point is \((1, 1)\).

To find the gradient of the curve at the point \((2, 1)\), we need to differentiate the equation implicitly with respect to \(x\).

The given equation is \(2x^2 - 4xy + 3y^2 = 3\).

Differentiate each term:

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

\(\frac{d}{dx}(-4xy) = -4y - 4x\frac{dy}{dx}\)

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

Substitute these into the differentiated equation:

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

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

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

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

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

Substitute \(x = 2\) and \(y = 1\):

\(\frac{dy}{dx} = \frac{4(1) - 4(2)}{6(1) - 4(2)} = \frac{4 - 8}{6 - 8} = \frac{-4}{-2} = 2\)

Thus, the gradient of the curve at the point \((2, 1)\) is 2.

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

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

Using the product rule, \(\frac{d}{dx}(x^2y) = 2xy + x^2\frac{dy}{dx}\).

For \(\frac{d}{dx}(ay^2) = 2ay\frac{dy}{dx}\).

Substitute these into the equation:

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

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

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

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

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

(b) For the tangent to be parallel to the y-axis, \(\frac{dy}{dx}\) is undefined, which occurs when the denominator is zero:

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

\(x^2 = 2ay\).

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

\(2ay^2 - ay^2 = 4a^3\).

\(ay^2 = 4a^3\).

\(y^2 = 4a^2\).

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

For \(y = 2a\), \(x^2 = 2a(2a) = 4a^2\), so \(x = 2a\) or \(x = -2a\).

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

(i) Differentiate both sides of \(\sqrt{x} + \sqrt{y} = \sqrt{a}\) with respect to \(x\):

\(\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0\)

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

\(\frac{1}{2\sqrt{y}} \frac{dy}{dx} = -\frac{1}{2\sqrt{x}}\)

\(\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}\)

(ii) At the intersection point \(P\), \(y = x\), so substitute into the curve equation:

\(\sqrt{x} + \sqrt{x} = \sqrt{a}\)

\(2\sqrt{x} = \sqrt{a}\)

\(\sqrt{x} = \frac{1}{2}\sqrt{a}\)

\(x = \frac{1}{4}a\)

Thus, \(P\) is \(\left( \frac{1}{4}a, \frac{1}{4}a \right)\).

The slope of the tangent at \(P\) is \(-\frac{\sqrt{y}}{\sqrt{x}} = -1\).

The equation of the tangent line is:

\(y - \frac{1}{4}a = -1(x - \frac{1}{4}a)\)

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

Rearranging gives:

\(x + y = \frac{1}{2}a\)

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

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

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

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

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

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

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

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

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

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

So, \(x = 0\) or \(x + 2y = 0\).

If \(x = 0\), substitute into the original equation:

\(3x^2y - y^3 = 3\) becomes \(-y^3 = 3\).

\(y = -\sqrt[3]{3}\).

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

If \(x + 2y = 0\), then \(y = -\frac{x}{2}\).

Substitute into the original equation:

\(x^3 + 3x^2(-\frac{x}{2}) - (-\frac{x}{2})^3 = 3\).

\(x^3 - \frac{3x^3}{2} - \frac{x^3}{8} = 3\).

\(-\frac{8x^3}{8} - \frac{12x^3}{8} - \frac{x^3}{8} = 3\).

\(-\frac{21x^3}{8} = 3\).

\(x^3 = -\frac{24}{21}\).

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

Point: \((-2, 1)\).