Step 1: Recall the inverse-derivative rule.
If $y=f(x)$ then $(f^{-1})'(y)=\frac{1}{f'(x)}$. So we need the $x$ with $f(x)=2$, then $f'$ there.
Step 2: Solve $f(x)=2$.
$\frac{3x^2+4x+1}{x^2+3x+2}=2$ gives $3x^2+4x+1=2x^2+6x+4$, so $x^2-2x-3=0$, that is $(x-3)(x+1)=0$.
Step 3: Pick the valid root.
The domain is $[0,\infty)$, so $x=3$ (we discard $x=-1$). Thus $f^{-1}(2)=3$.
Step 4: Factor before differentiating to save work.
Note $x^2+3x+2=(x+1)(x+2)$ and $3x^2+4x+1=(3x+1)(x+1)$, so $f(x)=\frac{3x+1}{x+2}$ after cancelling $(x+1)$. Much simpler.
Step 5: Differentiate the simplified form.
$f'(x)=\frac{3(x+2)-(3x+1)}{(x+2)^2}=\frac{5}{(x+2)^2}$. At $x=3$: $f'(3)=\frac{5}{25}=\frac{1}{5}$.
Step 6: Invert.
$(f^{-1})'(2)=\frac{1}{f'(3)}=\frac{1}{1/5}=5$, option 2.
\[ \boxed{5} \]