Sorry, but the answer is wrong. Thank you for trying though! This is the explanation for the correct answer:

The vertical asymptote of \(f(x)\) is \(x=2.\) Hence, \(d=2.\)

By long division,\(f(x) = \frac{1}{2} x - 2 - \frac{2}{2x - 4}.\)

Thus, the oblique asymptote of \(f(x)\) is \(y = \frac{1}{2} x - 2,\) which passes through \((0,-2).\) Therefore, the oblique asymptote of \(g(x)\) is \(y = -2x - 2.\)

Therefore, \(g(x) = -2x - 2 + \frac{k}{x - 2}\) for some constant \(k.\)

Finally, \(f(-2) = \frac{(-2)^2 - 6(-2) + 6}{2(-6) - 4} = -\frac{11}{4},\) so \(g(-2) = -2(-2) - 2 + \frac{k}{-2 - 2} = -\frac{11}{4}.\)

Solving, we find \(k = 19.\) Hence, \(g(x) = -2x - 2 + \frac{19}{x - 2} = \frac{-2x^2 + 2x + 23}{x - 2}.\)

We want to solve \(\frac{x^2 - 6x + 6}{2x - 4} = \frac{-2x^2 + 2x + 23}{x - 2}.\)

Then \(x^2 - 6x + 6 = -4x^2 + 4x + 46,\) or \(5x^2 - 10x - 40 = 0.\) This factors as \(5(x + 2)(x - 4) = 0,\) so the other point of intersection occurs at \(x = 4.\) Since \(f(4) = \frac{4^2 - 6 \cdot 4 + 6}{2(4) - 4} = -\frac{1}{2},\) the other point of intersection is \(\boxed{\left( 4, -\frac{1}{2} \right)}.\)

Still, thank you for trying, Guest!