(a) Differentiate the equation \(x^3 + 3xy^2 - y^3 = 5\) implicitly with respect to \(x\):

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

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

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

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

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

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

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

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

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

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

\(x^3 = 5\), so \(x = \sqrt[3]{5}\).

Point: \(\left( \sqrt[3]{5}, 0 \right)\).

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

\(x^3 + 3x(2x)^2 - (2x)^3 = 5\).

\(x^3 + 12x^3 - 8x^3 = 5\).

\(5x^3 = 5\), so \(x = 1\).

Then \(y = 2 \times 1 = 2\).

Point: (1, 2).