Question 547282
To find f(h(x)), we need to plug in h(x) for any x we see in f(x).  This leaves:

{{{f(x)= (3(6x+2)^2-6)/((6x+2)^2-4(6x+2))}}}
We could expand this... though it is not needed and will take some time.  Instead, we know that there is no limitations on x in the numerator.  But we do know that when there is a fraction, the denominator cannot be equal to 0.  
This means that 
{{{(6x+2)^2-4(6x+2)}}}  cannot equal 0.  

Well... when does this expression equal 0.  In other words, lets solve:
{{{(6x+2)^2-4(6x+2)=0}}} 
Again, we can expand this, or we could see that each term has a common factor of (6x+2) and factor this out. This leaves: 
{{{(6x+2)((6x+2)-4)=0}}} 
{{{(6x+2)(6x-2)=0}}}

If two things multiply to give a product of 0, then at least one of them must be 0.  This means 
{{{(6x+2)=0}}}
{{{6x=-2}}}
{{{x=-1/3}}}
or
{{{(6x-2)=0}}}
{{{6x=2}}}
{{{x=1/3}}}

This means x can be anything except {{{1/3}}} and {{{-1/3}}}.

Hopefully this helps!  Please let me know if you still do not understand:)