So lets say we have a space curve $r=\langle\cos t,\sin t,3t\rangle$ and we need to find the curvature at the point $P(1,0,0)$
$r'(t)=\langle-\sin t,\cos t,3\rangle$
$r''(t)=\langle-\cos t,-\sin t,0\rangle$
$||r'(t)||=\sqrt{(-\sin t)^2+(\cos t)^2+(3)^2} = \sqrt{1+9} = \sqrt{10}$
$||r'(t)||^3=(\sqrt{10})^3=10\sqrt{10}$
$r'(t) \times r''(t)$ = $\langle3\sin t,-3\cos t,1\rangle$
$||r'(t) \times r''(t)|| = \sqrt{(3\sin t)^2+(-3\cos t)^2+1^2}=\sqrt{10}$
$k=\frac{||r(t) \times r''(t)||}{||r'(t)||^3}=\frac{\sqrt{10}}{10\sqrt{10}}=\frac{1}{10}$
Is this correct? I think that's all you need to do for the problem, right?
Looks good to me. Another way is to reparametrize by arclength and use the simpler curvature formula.
Let $r(t) = \langle \cos(t),\sin(t),3t \rangle$. Then the arclength is written: $$ s(t) = \int_0^t ||r'(\tau)||_2d\tau=\sqrt{10}t \;\;\;\implies\;\;\;t = \frac{s}{\sqrt{10}} $$ So, using this parametrization, we get: \begin{align} r(s) &= \langle \cos(s/\sqrt{10}),\sin(s/\sqrt{10}),3t/\sqrt{10} \rangle\\ r'(s) &= \langle -\sin(s/\sqrt{10})/\sqrt{10},\cos(s/\sqrt{10})/\sqrt{10},3/\sqrt{10} \rangle\\ r''(s) &= \langle -\cos(s/\sqrt{10})/{10},-\sin(s/\sqrt{10})/{10},0 \rangle \end{align} The curvature is then computable by a simpler formula, i.e: $$ \kappa(s) = ||r''(s)||_2=\sqrt{(-\cos(s/\sqrt{10})/{10})^2 + (-\sin(s/\sqrt{10})/{10}})^2 = \frac{1}{10} $$ as you computed as well.