Answer
The inverse function is:
$$f^{-1}(x)=\frac{(x-1)^2-2}{3}.$$
Work Step by Step
For the function $f$ to be defined $2+3x\geq0\Rightarrow 3x\geq-2\Rightarrow x\geq-2/3.$
Now let $y=f(x)=1+\sqrt{2+3x}$. We need to express $x$ in terms of $y$ because then $x=f^{-1}(y)$.
$$y=1+\sqrt{2+3x}\Rightarrow y-1=\sqrt{2+3x}$$
squaring the last equality
$$(y-1)^2=2+3x\Rightarrow 3x=(y-1)^2-2$$
and finally
$$x=\frac{(y-1)^2-2}{3}$$
which means
$$f^{-1}(y)=\frac{(y-1)^2-2}{3}.$$
Renaming $y$ back to $x$ we get
$$f^{-1}(x)=\frac{(x-1)^2-2}{3}.$$
