(a) The vertical asymptote occurs where the denominator is zero: \(x + 1 = 0 \Rightarrow x = -1\).
The oblique asymptote is found by dividing \(x^2 - x\) by \(x + 1\), giving \(y = x - 2\).

(b) Differentiate \(y = \frac{x^2 - x}{x + 1}\) using the quotient rule:
\(\frac{dy}{dx} = \frac{(2x - 1)(x + 1) - (x^2 - x)}{(x + 1)^2} = \frac{1 - 2(x + 1)^2}{(x + 1)^2}\).
Set \(\frac{dy}{dx} = 0\) to find stationary points:
\(1 - 2(x + 1)^2 = 0 \Rightarrow (x + 1)^2 = \frac{1}{2}\).
Solving gives \(x = -1 \pm \sqrt{2}\).
Substitute back to find \(y\):
\(y = \frac{(-1 + \sqrt{2})^2 - (-1 + \sqrt{2})}{-1 + \sqrt{2} + 1} = -3 + 2\sqrt{2}\),
\(y = \frac{(-1 - \sqrt{2})^2 - (-1 - \sqrt{2})}{-1 - \sqrt{2} + 1} = -3 - 2\sqrt{2}\).

(c) The sketch shows the curve with asymptotes and intersections at \((0, 0)\) and \((1, 0)\).

(d) Solve \(\left| \frac{x^2 - x}{x + 1} \right| < 6\):
\(\frac{x^2 - x}{x + 1} < 6\) or \(\frac{x^2 - x}{x + 1} > -6\).
For \(\frac{x^2 - x}{x + 1} < 6\):
\(x^2 - 7x - 6 = 0 \Rightarrow x = -3, -2\).
For \(\frac{x^2 - x}{x + 1} > -6\):
\(x^2 + 5x + 6 = 0 \Rightarrow x = -2, -3\).
Thus, \(-3 < x < -2 \text{ and } \frac{1}{2} - \frac{1}{2}\sqrt{73} < x < \frac{1}{2} + \frac{1}{2}\sqrt{73}\).