Let $y$ be an arbitrary element in range of $f$.
Let $y=9 x^2+6 x-5=9 x^2+6 x+1-6$
$$
\begin{aligned}
&\Rightarrow \mathrm{y}=(3 \mathrm{x}+1)^2-6 \\
&\Rightarrow \mathrm{y}+6=(3 \mathrm{x}+1)^2 \quad \Rightarrow 3 \mathrm{x}+1=\sqrt{\mathrm{y}+6} \\
&\Rightarrow \mathrm{x}=\frac{\sqrt{\mathrm{y}+6}-1}{3}=\mathrm{g}(\mathrm{y})
\end{aligned}
$$
Let $\mathrm{g}:$ Range $\mathrm{f} \rightarrow \mathrm{R}^{+}$, such that $\mathrm{g}(\mathrm{y})=\frac{\sqrt{\mathrm{y}+6}-1}{3}$ $g o f(x)=g(f(x))=g\left(9 x^2+6 x-5\right)=g\left[(3 x+1)^2-6\right]$
$$
=\frac{\left(\sqrt{(3 x+1)^2-6+6}\right)-1}{3}=\frac{3 x+1-1}{3}=x
$$
$$
\Rightarrow \mathrm{gof}(\mathrm{x})=\mathrm{x}
$$
Now fog $(y)=f(g(y))=f\left(\frac{\sqrt{y+6}-1}{3}\right)$
$$
\begin{aligned}
&=\left[3\left(\frac{\sqrt{\mathrm{y}+6}-1}{3}\right)+1\right]^2-6=\mathrm{y} \\
\Rightarrow & \text { fog }(\mathrm{y})=\mathrm{y} \Rightarrow \text { gof }=\mathrm{I}_{\mathrm{x}}, \text { fog }=\mathrm{I}_{\mathrm{y}} \\
\Rightarrow & \mathrm{f} \text { is invertible } \mathrm{f}^{-1}(\mathrm{y})=\mathrm{g}(\mathrm{y})=\frac{\sqrt{\mathrm{y}+6}-1}{3}
\end{aligned}
$$