Answer
$$\kappa \left( t \right) = \frac{{6t - 6{t^2}}}{{{{\left( {9{t^4} + 4{t^2} - 4t + 1} \right)}^{3/2}}}}$$
Work Step by Step
$$\eqalign{
& x = 1 - {t^3},\,\,\,\,y = t - {t^2} \cr
& {\text{Let }}{\bf{r}}\left( t \right) = x\left( t \right){\bf{i}} + y\left( t \right){\bf{j}} \cr
& {\text{Then}} \cr
& {\bf{r}}\left( t \right) = \left( {1 - {t^3}} \right){\bf{i}} + \left( {t - {t^2}} \right){\bf{j}} \cr
& {\text{Calculate the derivatives }}{\bf{r}}'\left( t \right){\text{ and }}{\bf{r}}{\text{''}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {1 - {t^3}} \right]{\bf{i}} + \frac{d}{{dt}}\left[ {t - {t^2}} \right]{\bf{j}} \cr
& {\bf{r}}'\left( t \right) = - 3{t^2}{\bf{i}} + \left( {1 - 2t} \right){\bf{j}} \cr
& and \cr
& {\bf{r}}''\left( t \right) = \frac{d}{{dt}}\left[ { - 3{t^2}} \right]{\bf{i}} + \frac{d}{{dt}}\left[ {1 - 2t} \right]{\bf{j}} \cr
& {\bf{r}}''\left( t \right) = - 6t{\bf{i}} - 2{\bf{j}} \cr
& \cr
& {\text{Calculate the cross product }}{\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right) \cr} $$
\[\begin{gathered}
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
{ - 3{t^2}}&{1 - 2t}&0 \\
{ - 6t}&{ - 2}&0
\end{array}} \right| \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{1 - 2t}&0 \\
{1 - 2t}&0
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
{ - 3{t^2}}&0 \\
{9{e^{ - 6t}}}&0
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
{ - 3{t^2}}&{1 - 2t} \\
{ - 6t}&{ - 2}
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = 0{\mathbf{i}} - 0{\mathbf{j}} + \left[ { - 3{t^2}\left( { - 2} \right) - \left( {1 - 2t} \right)\left( { - 6t} \right)} \right]{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left( {6{t^2} + 6t - 12{t^2}} \right){\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left( {6t - 6{t^2}} \right){\mathbf{k}} \hfill \\
\end{gathered} \]
$$\eqalign{
& {\text{Use the formula }}\left( 3 \right)\,\,\,\kappa \left( t \right) = \frac{{\left\| {{\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right)} \right\|}}{{{{\left\| {{\bf{r}}'\left( t \right)} \right\|}^3}}}.{\text{ Thus}}{\text{,}} \cr
& \kappa \left( t \right) = \frac{{\left\| {\left( {6t - 6{t^2}} \right){\bf{k}}} \right\|}}{{{{\left\| { - 3{t^2}{\bf{i}} + \left( {1 - 2t} \right){\bf{j}}} \right\|}^3}}} \cr
& \kappa \left( t \right) = \frac{{6t - 6{t^2}}}{{{{\left( {\sqrt {{{\left( { - 3{t^2}} \right)}^2} + {{\left( {1 - 2t} \right)}^2}} } \right)}^3}}} \cr
& \kappa \left( t \right) = \frac{{6t - 6{t^2}}}{{{{\left( {\sqrt {9{t^4} + 4{t^2} - 4t + 1} } \right)}^3}}} \cr
& \kappa \left( t \right) = \frac{{6t - 6{t^2}}}{{{{\left( {9{t^4} + 4{t^2} - 4t + 1} \right)}^{3/2}}}} \cr} $$
