Answer
\[
-\frac{\sqrt{2} e^{2} t}{\left(1+e^{2} t\right)^{2}}=\tau(t)
\]
Work Step by Step
Using the formula
\[
\frac{[\overrightarrow{r^{3}}(t) \times \overrightarrow{r^{\prime \prime}}(t)] \cdot \overrightarrow{r^{\prime \prime \prime}}(t)}{\|\overrightarrow{r^{\prime}}(t) \times \overrightarrow{r^{\prime \prime}}(t)\|^{2}}=\tau(t)
\]
For the curve $e^{t \hat{i}}+e^{-t \hat{i}}+\sqrt{2} t \hat{k}=\vec{r}(t)$
\[
\begin{aligned}
\vec{r}(t)=\frac{d \vec{r}(t)}{d t} &=e^{t \hat{t}}+e^{-t} \hat{i}+\sqrt{2} t \hat{k} \\
\overrightarrow{r^{\prime \prime}}(t) &=\frac{\overrightarrow{d r^{\prime}}(t)}{d t}=e^{t} \hat{i}+e^{-t \hat{i}} \\
\overrightarrow{r^{\prime \prime \prime}}(t) &=\frac{d \overrightarrow{r^{\prime \prime}}(t)}{d t}=e^{t \hat{i}}-e^{-t \hat{i}}
\end{aligned}
\]
So
\[
\vec{r}(t) \times \overrightarrow{r^{\prime \prime}}(t)=\| \begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
e^{t} & e^{-t} & \sqrt{2} \\
e^{t} & e^{-t} & 0
\end{array}||=-\sqrt{2} e^{-t} \hat{i}+\sqrt{2} e^{t} \hat{j}+2 \hat{k}
\]
\[
\|\overrightarrow{r^{\prime}}(t) \times \overrightarrow{r^{\prime \prime}}(t)\|^{2}=2 e^{-2 t}+2 e^{2} t+4=\frac{2}{e^{2 t}}\left(e^{4 l}+1+2 e^{2} t\right)=\frac{2}{e^{2 t}}\left(1+e^{2 t}\right)^{2}
\]
Using the results:
\[
\begin{aligned}
\tau(t)=& \frac{\left[-\sqrt{2} e^{-t} \hat{i}+\sqrt{2} e^{t} \hat{j}+2 \hat{k}\right] \cdot\left[e^{t} \hat{i}-e^{-t} \hat{j}\right]}{\frac{2}{e^{2 t} t}\left(e^{2 t}+1\right)^{2}} \\
&=\frac{\sqrt{2}|-1-1| c^{2} t}{2\left(c^{2} t+1\right)^{2}}=-\frac{\sqrt{2} e^{2} t}{\left(e^{2} t+1\right)^{2}}
\end{aligned}
\]
\[
-\frac{\sqrt{2} e^{2} t}{\left(1+e^{2} t\right)^{2}}=\tau(t)
\]
