$$\gamma(t)=(\log t\sqrt2,\frac{1}{3t},3t+1)$$ $$t\in [1/3,3]$$
my try :
the tangent to the curve is : $$\gamma'(t)=\left(\frac{\sqrt2}{t},\frac{-1}{3t^2},3\right)$$
the norm of the tangent to the curve: $$||\gamma(t)||=\left(\frac{2}{t^2}+\frac{1}{9t^4}+9\right)^\frac{1}{2}$$
expanding and simplifying : $$||\gamma(t)||=\frac{((12t^2+5)(12t^2+1))^\frac{1}{2}}{6t^2}$$
I don't know how to proceed to solve the integral from this
I am assuming the curve is,
$$ \displaystyle ~\gamma(t)=(\sqrt2 \ln t,\frac{1}{3t},3t+1), t \in \left[1/3, 3\right]$$
Your work is correct till the step,
$ \displaystyle ||\gamma(t)|| = \sqrt{\frac{2}{t^2}+\frac{1}{9t^4}+9}$
But your simplification is incorrect.
$ \displaystyle ||\gamma(t)|| = \sqrt{\frac{2}{t^2}+\frac{1}{9t^4}+9} = \sqrt{\left(\frac 1 {3 t^2} + 3 \right)^2}$
So length of the curve is,
$ \displaystyle \int_{1/3}^3 \left(\frac 1 {3 t^2} + 3 \right) ~ dt$
which is a straightforward integral.