The question is to find the curvature of the curve $r(t)=\langle t^2, \ln{t}, t\ln{t}\rangle$ at point $(1, 0, 0).$
I've found $r^{\prime} = \langle 2t, 1/t, \ln{t} + 1 \rangle$ and $r^{\prime\prime} = \langle 2, -t^{-2}, 1/t \rangle$ and got $$\lvert r^{\prime}\rvert =\sqrt{4t^2 +1/t^2 +\ln^{2}{t} + 2\ln{t} + 1}$$
The cross product I got use in for $$\frac{|r^{\prime}\times r^{\prime\prime}|}{|r^{\prime}|}$$ wasn't much less of a complex mess to deal with: $$(1/4 - ( \ln{t} + 1)(1/2\sqrt{t}))\mathbf{i} - (t-2\ln{t}-2)\mathbf{j} + (2t/2\sqrt{t}-1)\mathbf{k}$$
Am I doing these right? I don't see anything you could do with all this. The answer is supposed to be $\frac{1}{7}\sqrt{\frac{19}{14}}.$
Starting from $r(t) = (t^{2},\ln{t},t\ln{t}),$ we have $$r^{\prime}(t) = (2t,1/t,1+\ln{t})\quad\text{and}\quad r^{\prime\prime}(t) = (2,-1/t^{2},1/t).$$
Hence, $$r^{\prime}(1) = (2,1,1)\quad\text{and}\quad r^{\prime\prime}(1) = (2,-1,1).$$ Now we have $$\lvert r^{\prime}(1)\times r^{\prime\prime}(1) \rvert = \lvert (2,0,-4) \rvert = \sqrt{20}$$ and $$\lvert r^{\prime}(1) \rvert = \lvert (2,1,1) \rvert = \sqrt{6}.$$ Therefore the curvature at $(1,0,0)$ appears to be $$\frac{\lvert r^{\prime}(1)\times r^{\prime\prime}(1) \rvert}{\lvert r^{\prime}(1) \rvert^{3}} = \frac{\sqrt{30}}{18}.$$