Question 206612
{{{1/3x+9 = x+1/x^2+3x + 1/3x^2+9x}}}  is what you wrote.  I think that when you changed your fractions from vertical to horizontal you forgot the parentheses, so I think this is what you meant (correct me if I'm wrong):
{{{1/(3x+9) = (x+1)/(x^2+3x) + 1/(3x^2+9x)}}}   Factor everything you can.
{{{ 1/(3(x+3)) = (x+1)/(x(x+3)) + 1/(3x(x+3)) }}}  
Get a (least) common denominator of all the fractions which is 3x(x+3)  (why is that the LCD?)
{{{ x/(3x(x+3)) = 3(x+1)/(3x(x+3)) + 1/(3x(x+3)) }}}  and multiply both sides by the LCD to get rid of fractions.
{{{x = 3x+3 + 1}}}  subtract 3x from both sides and combine like terms
{{{-2x = 4}}}
{{{x = -2}}}
 
check answer in ORIGINAL
{{{1/(3(-2)+9) = ((-2)+1)/((-2)^2+3(-2)) + 1/(3(-2)^2+9(-2))}}} ?
{{{1/3 = (-1)/(4-6) + 1/(12-18)}}} ?
{{{1/3 = 1/2 - 1/6=(3-1)/6=1/3}}} Yes!