To find the gradient of the curve \(x^3 + 3xy^2 - y^3 = 1\), we need to differentiate implicitly with respect to \(x\).

The derivative of \(x^3\) is \(3x^2\).

For \(3xy^2\), use the product rule: \(3(y^2 + 2xy\frac{dy}{dx}) = 3y^2 + 6xy\frac{dy}{dx}\).

The derivative of \(y^3\) is \(3y^2\frac{dy}{dx}\).

Substituting these into the equation, we get:

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

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

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

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

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

Substitute \(x = 1\) and \(y = 3\):

\(\frac{dy}{dx} = \frac{-3(1)^2 - 3(3)^2}{6(1)(3) - 3(3)^2}\).

\(\frac{dy}{dx} = \frac{-3 - 27}{18 - 27}\).

\(\frac{dy}{dx} = \frac{-30}{-9} = \frac{10}{3}\).