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