You also need to convert the $\displaystyle \begin{align*} x^3 \end{align*}$ from your trigonometric substitution.
Since you have edited your original post I can't really state what we originally had. Anyway, you have correctly gotten to $\displaystyle \begin{align*} S = \int_0^3{ 2\pi\left( x^3 \right) \, \sqrt{ 1 + \left( 3x^2 \right) ^2 } \, \mathrm{d}x} \end{align*}$. From here I would do...
$\displaystyle \begin{align*} \int_0^3{ 2\pi \left( x^3 \right) \, \sqrt{ 1 + \left( 3x^2 \right) ^2 } \, \mathrm{d}x } &= 2\pi \int_0^3{ x^3 \, \sqrt{ 1 + \left( 3x^2 \right) ^2 } \, \mathrm{d}x } \\ &= \frac{\pi}{3} \int_0^3{ x^2 \, \sqrt{1 + \left( 3x^2 \right) ^2 } \, 6x \, \mathrm{d}x } \\ &= \frac{\pi}{9} \int_0^3{ 3x^2 \, \sqrt{ 1 + \left( 3x^2 \right) ^2 } \, 6x \, \mathrm{d}x } \end{align*}$
Now I would substitute $\displaystyle \begin{align*} 3x^2 = \tan{ \left( \theta \right) } \implies 6x\,\mathrm{d}x = \sec^2{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}$, and also note that when $\displaystyle \begin{align*} x = 0, \theta = 0 \end{align*}$ and when $\displaystyle \begin{align*} x = 3, \theta = \arctan{ \left( 27 \right) } \end{align*}$, the integral becomes
$\displaystyle \begin{align*} \frac{\pi}{9} \int_0^3{ 3x^2 \, \sqrt{ 1 + \left( 3x^2 \right) ^2 } \, 6x \, \mathrm{d}x } &= \frac{\pi}{9} \int_0^{\arctan{ \left( 27 \right) } } { \tan{ \left( \theta \right) } \, \sqrt{ 1 + \tan^2{ \left( \theta \right) } } \, \sec^2{ \left( \theta \right) } \, \mathrm{d}\theta } \end{align*}$
Can you go from here?
This problem needs u substitution, not trig sub. It's just like an arc length problem before. Same numbers.
evaluated at upper bound 3 and lower bound 0
evaluated at upper bound 3 and lower bound 0