jmc

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 $\boxed{\left(4,-\frac{1}{2}\right)}.$