(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = 7 - x^2 - 6x\):

\(y = \int (7 - x^2 - 6x) \, dx = 7x - \frac{x^3}{3} - \frac{6x^2}{2} + c\).

Using the point \((3, -10)\), substitute to find \(c\):

\(-10 = 7(3) - \frac{3^3}{3} - \frac{6(3)^2}{2} + c\)

\(-10 = 21 - 9 - 27 + c\)

\(c = 5\)

Thus, the equation is \(y = 7x - \frac{x^3}{3} - \frac{6x^2}{2} + 5\).

(ii) Express \(7 - x^2 - 6x\) in the form \(a - (x + b)^2\):

Complete the square for \(-x^2 - 6x\):

\(-x^2 - 6x = -(x^2 + 6x) = -(x + 3)^2 + 9\)

Thus, \(7 - x^2 - 6x = 7 + 9 - (x + 3)^2 = 16 - (x + 3)^2\)

So, \(a = 16\) and \(b = 3\).

(iii) Find the set of values of \(x\) for which the gradient is positive:

\(7 - x^2 - 6x > 0\)

Using the completed square form: \(16 - (x + 3)^2 > 0\)

\((x + 3)^2 < 16\)

\(-4 < x + 3 < 4\)

\(-7 < x < 1\)