Solution

We know that

$(f^{-1})'(y)=\frac{1}{f'(x)}$, where $f(x)=y$

Here, we need $(f^{-1})'(2)$.

So first find $x$ such that

$f(x)=2$

$\frac{3x^2+4x+1}{x^2+3x+2}=2$

$3x^2+4x+1=2x^2+6x+4$

$x^2-2x-3=0$

$(x-3)(x+1)=0$

Since domain is $[0,\infty)$,

$x=3$

Now,

$f(x)=\frac{3x^2+4x+1}{x^2+3x+2}$

Let

$N=3x^2+4x+1$

and

$D=x^2+3x+2$

Then,

$f'(x)=\frac{N'D-ND'}{D^2}$

Now at $x=3$,

$N=3(3)^2+4(3)+1=40$

$D=(3)^2+3(3)+2=20$

$N'=6x+4$

So,

$N'=22$

$D'=2x+3$

So,

$D'=9$

Therefore,

$f'(3)=\frac{22\cdot 20-40\cdot 9}{20^2}$

$f'(3)=\frac{440-360}{400}$

$f'(3)=\frac{80}{400}$

$f'(3)=\frac{1}{5}$

Hence,

$(f^{-1})'(2)=\frac{1}{f'(3)}$

$=5$