Answer
$$\kappa \left( t \right) = \frac{{12{e^{2t}}}}{{{{\left( {9{e^{6t}} + {e^{ - 2t}}} \right)}^{3/2}}}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = {e^{3t}}{\bf{i}} + {e^{ - t}}{\bf{j}} \cr
& {\text{Calculate the derivatives }}{\bf{r}}'\left( t \right){\text{ and }}{\bf{r}}{\text{''}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {{e^{3t}}} \right]{\bf{i}} + \frac{d}{{dt}}\left[ {{e^{ - t}}} \right]{\bf{j}} \cr
& {\bf{r}}'\left( t \right) = 3{e^{3t}}{\bf{i}} - {e^{ - t}}{\bf{j}} \cr
& and \cr
& {\bf{r}}''\left( t \right) = \frac{d}{{dt}}\left[ {3{e^{3t}}} \right]{\bf{i}} + \frac{d}{{dt}}\left[ { - {e^{ - t}}} \right]{\bf{j}} \cr
& {\bf{r}}''\left( t \right) = 9{e^{3t}}{\bf{i}} + {e^{ - t}}{\bf{j}} \cr
& \cr
& {\text{Calculate the cross product }}{\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right) \cr} $$
\[\begin{gathered}
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
{3{e^{3t}}}&{ - {e^{ - t}}}&0 \\
{9{e^{3t}}}&{{e^{ - t}}}&0
\end{array}} \right| \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{ - {e^{ - t}}}&0 \\
{{e^{ - t}}}&0
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
{3{e^{3t}}}&0 \\
{9{e^{3t}}}&0
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
{3{e^{3t}}}&{ - {e^{ - t}}} \\
{9{e^{3t}}}&{{e^{ - t}}}
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = 0{\mathbf{i}} - 0{\mathbf{j}} + \left( {3{e^{2t}} + 9{e^{2t}}} \right){\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = 12{e^{2t}}{\mathbf{k}} \hfill \\
\end{gathered} \]
$$\eqalign{
& {\text{Use the formula }}\left( 3 \right)\,\,\,\kappa \left( t \right) = \frac{{\left\| {{\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right)} \right\|}}{{{{\left\| {{\bf{r}}'\left( t \right)} \right\|}^3}}}.{\text{ Thus}}{\text{,}} \cr
& \kappa \left( t \right) = \frac{{\left\| {12{e^{2t}}{\bf{k}}} \right\|}}{{{{\left\| {3{e^{3t}}{\bf{i}} - {e^{ - t}}{\bf{j}}} \right\|}^3}}} \cr
& \kappa \left( t \right) = \frac{{12{e^{2t}}}}{{{{\left( {\sqrt {{{\left( {3{e^{3t}}} \right)}^2} + {{\left( { - {e^{ - t}}} \right)}^2}} } \right)}^3}}} \cr
& \kappa \left( t \right) = \frac{{12{e^{2t}}}}{{{{\left( {\sqrt {9{e^{6t}} + {e^{ - 2t}}} } \right)}^3}}} \cr
& \kappa \left( t \right) = \frac{{12{e^{2t}}}}{{{{\left( {9{e^{6t}} + {e^{ - 2t}}} \right)}^{3/2}}}} \cr} $$
