Given \(\frac{dy}{dx} = 3x^2 + ax + b\), stationary points occur where \(\frac{dy}{dx} = 0\).

For \(x = -1\), \(3(-1)^2 + a(-1) + b = 0\) gives:

\(3 - a + b = 0\)

For \(x = 3\), \(3(3)^2 + a(3) + b = 0\) gives:

\(27 + 3a + b = 0\)

Solving these equations simultaneously:

1. \(3 - a + b = 0\)

2. \(27 + 3a + b = 0\)

Subtract equation 1 from equation 2:

\(27 + 3a + b - (3 - a + b) = 0\)

\(27 + 3a + b - 3 + a - b = 0\)

\(24 + 4a = 0\)

\(a = -6\)

Substitute \(a = -6\) into equation 1:

\(3 - (-6) + b = 0\)

\(3 + 6 + b = 0\)

\(b = -9\)

Integrate \(\frac{dy}{dx} = 3x^2 - 6x - 9\) to find \(y\):

\(y = x^3 - 3x^2 - 9x + c\)

Use point \((-1, 2)\):

\(2 = (-1)^3 - 3(-1)^2 - 9(-1) + c\)

\(2 = -1 - 3 + 9 + c\)

\(c = -3\)

Use point \((3, k)\):

\(k = 3^3 - 3(3)^2 - 9(3) - 3\)

\(k = 27 - 27 - 27 - 3\)

\(k = -30\)