We start with the second derivative \(\frac{d^2y}{dx^2} = 2x - 1\). To find the first derivative \(\frac{dy}{dx}\), integrate:

\(\frac{dy}{dx} = \int (2x - 1) \, dx = x^2 - x + c\).

Given the minimum point at \(x = 3\), \(\frac{dy}{dx} = 0\) at this point:

\(0 = 3^2 - 3 + c\) which simplifies to \(c = -6\).

Thus, \(\frac{dy}{dx} = x^2 - x - 6\).

Setting \(\frac{dy}{dx} = 0\) to find critical points:

\(x^2 - x - 6 = 0\).

Factoring gives \((x - 3)(x + 2) = 0\), so \(x = 3\) or \(x = -2\).

Since \(x = 3\) is a minimum, \(x = -2\) is a maximum.

Integrate \(\frac{dy}{dx}\) to find \(y\):

\(y = \int (x^2 - x - 6) \, dx = \frac{1}{3}x^3 - \frac{1}{2}x^2 - 6x + k\).

Given \(y = -10\) when \(x = 3\), substitute to find \(k\):

\(-10 = \frac{1}{3}(3)^3 - \frac{1}{2}(3)^2 - 6(3) + k\).

\(-10 = 9 - 4.5 - 18 + k\).

\(k = \frac{3}{2}\).

Thus, \(y = \frac{1}{3}x^3 - \frac{1}{2}x^2 - 6x + \frac{3}{2}\).

Substitute \(x = -2\) to find \(y\) at the maximum point:

\(y = \frac{1}{3}(-2)^3 - \frac{1}{2}(-2)^2 - 6(-2) + \frac{3}{2}\).

\(y = -\frac{8}{3} - 2 + 12 + \frac{3}{2}\).

\(y = 10.8\).

Therefore, the coordinates of the maximum point are \((-2, 10.8)\).