ok, so f'(x)=12x^2-4x+5

Now you want to plug in x=6 to find the slope of the tangent line:

12(6)^2-4(6)+5

=432-24+5=413

Now we're going to find the y coordinate when x=6

f(x)=4x^3-2x^2+5x-3

y=4(6)^3-2(6)^2+5(6)-3=819

so your point is (6,819)

Now use that to find the equation for the tangent line:

y-819=413(x-6)

y-819=413x-2487

y=413x-1659

I'm pretty sure that's right. If anybody sees something I missed feel free to comment.