A little bit of simplification first, then yes, a trigonometric substitution

$\displaystyle \begin{align*} S &= 2\pi \int_2^3{ y^3 \, \sqrt{ 1 + \left( 3y^2 \right) ^2 } \, \mathrm{d}y } \\ &= \frac{\pi}{3} \int_2^3{ y^2 \, \sqrt{ 1 + \left( 3y ^2 \right) ^2 } \, 6y\, \mathrm{d}y } \\ &= \frac{\pi}{9} \int_2^3{ 3y^2 \, \sqrt{ 1 + \left( 3y^2 \right) ^2 } \, 6y \, \mathrm{d}y } \end{align*}$

So now substitute $\displaystyle \begin{align*} 3y^2 = \tan{ \left( \theta \right) } \implies 6y \, \mathrm{d}y = \sec^2{ \left( \theta \right) } \, \mathrm{d}\theta \end{align*}$ and notice that when $\displaystyle \begin{align*} y = 2, \theta = \arctan{ \left( 12 \right) } \end{align*}$ and when $\displaystyle \begin{align*} y = 3, \theta = \arctan{ \left( 27 \right) } \end{align*}$, the integral becomes

$\displaystyle \begin{align*} \frac{\pi}{9} \int_2^3{ 3y^2 \, \sqrt{ 1 + \left( 3y^2 \right) ^2 } \, 6y\, \mathrm{d}y } &= \frac{\pi}{9} \int_{\arctan{ \left( 12 \right) } } ^{\arctan{ \left( 27 \right) } } { \tan{ \left( \theta \right) } \, \sqrt{ 1 + \tan^2{ \left( \theta \right) } } \, \sec^2{ \left( \theta \right) } \, \mathrm{d}\theta } \end{align*}$

Go from here.