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