Given $f\left( x \right)=3x+5$

$\implies f'\left( x \right)=3>0$

$\implies f$ is strictly increasing function.

$\implies f$ is one-one function.

Also, $R_f=R=$ co-domain

$\implies f$ is onto function.

Thus, $f$ is a bijective function.

Hence, $f^{-1}$ exists.

Let $y=3x+5$

$\implies x= \frac{y-5}{3}$

Thus, $f^{-1}(x)=\frac{x-5}{3}$.