I began with the formula: $k(t) = \frac{||\mathbf r'(t) \times\mathbf r''(t)|| }{ ||\mathbf r'(t)||^3}$.
My $\mathbf r'(t) = \langle -3t^{-2}, 0, 1\rangle$ and $||\mathbf r'(t)|| = \sqrt{9t^{-4}+1}$.
Thus, $\mathbf r''(t) = \langle 6t^{-3}, 0, 0\rangle$.
Taking the cross product of $\mathbf r' $ and $\mathbf r'' $ was $\langle 0, 6t^{-3}, 0\rangle$. The length of this is $6/t^3$.
When I divide this by $||\mathbf r'(t)||^3$, my homework says the answer is incorrect. What did I do wrong?
After the simplifications you have arrived at $k(t)=\frac{6/t^3}{(9/t^4+1)^{3/2}}$. We can multiply top and bottom by $t^6=(t^4)^{3/2}$ to remove the singularity at $t=0$: $$k(t)=\frac{6t^3}{(9+t^4)^{3/2}}$$ Try and see if this works.