Answer
$$\frac{{d{\bf{r}}}}{{d\tau }} = 2\tau {e^{{\tau ^2}}}{\bf{i}} - 8\tau {e^{ - {\tau ^2}}}{\bf{j}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}} = {e^t}{\bf{i}} + 4{e^{ - t}}{\bf{j}};{\text{ }}t = {\tau ^2} \cr
& {\text{Calculate }}\frac{{d{\bf{r}}}}{{d\tau }}{\text{using }}\frac{{d{\bf{r}}}}{{d\tau }} = \frac{{d{\bf{r}}}}{{dt}}\frac{{dt}}{{d\tau }} \cr
& {\text{ }}\frac{{d{\bf{r}}}}{{d\tau }} = \frac{d}{{dt}}\left[ {{e^t}{\bf{i}} + 4{e^{ - t}}{\bf{j}}} \right]\frac{d}{{d\tau }}\left[ {{\tau ^2}} \right] \cr
& {\text{ }}\frac{{d{\bf{r}}}}{{d\tau }} = \left( {{e^t}{\bf{i}} - 4{e^{ - t}}{\bf{j}}} \right)\left( {2\tau } \right) \cr
& {\text{ }}\frac{{d{\bf{r}}}}{{d\tau }} = 2\tau {e^t}{\bf{i}} - 8\tau {e^{ - t}}{\bf{j}} \cr
& {\text{Write in terms of }}\tau \cr
& {\text{ }}\frac{{d{\bf{r}}}}{{d\tau }} = 2\tau {e^{{\tau ^2}}}{\bf{i}} - 8\tau {e^{ - {\tau ^2}}}{\bf{j}} \cr
& \cr
& {\text{Expressing }}{\bf{r}}{\text{ in terms of }}\tau {\text{ and differentiate}} \cr
& {\bf{r}} = {e^{{\tau ^2}}}{\bf{i}} + 4{e^{ - {\tau ^2}}}{\bf{j}} \cr
& {\text{Calculate }}\frac{{d{\bf{r}}}}{{d\tau }} \cr
& \frac{{d{\bf{r}}}}{{d\tau }} = {\text{ }}\frac{d}{{d\tau }}\left[ {{e^{{\tau ^2}}}{\bf{i}} + 4{e^{ - {\tau ^2}}}{\bf{j}}} \right] \cr
& \frac{{d{\bf{r}}}}{{d\tau }} = {\text{ }}{e^{{\tau ^2}}}\left( {2\tau } \right){\bf{i}} + 4{e^{ - {\tau ^2}}}\left( { - 2\tau } \right){\bf{j}} \cr
& \frac{{d{\bf{r}}}}{{d\tau }} = 2\tau {e^{{\tau ^2}}}{\bf{i}} - 8\tau {e^{ - {\tau ^2}}}{\bf{j}} \cr} $$
