$\displaystyle f(x)=\frac{3}{2x+1}$
I got
the limit = $\displaystyle 3(\frac{-2}{ (2x+1)^2}) $
When you use the definition, the limit will involve (depending on the form you use), the letter h, a, x_0, or delta x.
You should have
$\displaystyle \lim_{x \to h}\frac{f(x + h) - f(x)}{h}$
$\displaystyle \lim_{x \to h}\frac{\frac{3}{2(x + h) + 1} - \frac{3}{2x + 1}}{h}$
for starters, at least.
For f(x) = 3/(2x + 1)
f'(x) = -6/(2x + 1)^2 , which is what you have (after simplifying).
You can check using wolfram... y = 3/(2x + 1) - Wolfram|Alpha