Answer
$${\text{ }}\frac{{d{\bf{r}}}}{{d\tau }} = - \frac{9}{{2{\tau ^{5/2}}}}{\bf{j}} - \frac{1}{{{\tau ^2}}}{\bf{k}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}} = {\bf{i}} + 3{t^{3/2}}{\bf{j}} + t{\bf{k}};{\text{ }}t = 1/\tau \cr
& {\text{Calculate }}\frac{{d{\bf{r}}}}{{d\tau }}{\text{using }}\frac{{d{\bf{r}}}}{{d\tau }} = \frac{{d{\bf{r}}}}{{dt}}\frac{{dt}}{{d\tau }} \cr
& {\text{ }}\frac{{d{\bf{r}}}}{{d\tau }} = \frac{d}{{dt}}\left[ {{\bf{i}} + 3{t^{3/2}}{\bf{j}} + t{\bf{k}}} \right]\frac{d}{{d\tau }}\left[ {\frac{1}{\tau }} \right] \cr
& {\text{ }}\frac{{d{\bf{r}}}}{{d\tau }} = \left( {\frac{9}{2}{t^{1/2}}{\bf{j}} + {\bf{k}}} \right)\left( { - \frac{1}{{{\tau ^2}}}} \right) \cr
& {\text{ }}\frac{{d{\bf{r}}}}{{d\tau }} = - \frac{9}{{2{\tau ^2}}}{t^{ - 3/2}}{\bf{j}} - \frac{1}{{{\tau ^2}}}{\bf{k}} \cr
& {\text{Write in terms of }}\tau \cr
& {\text{ }}\frac{{d{\bf{r}}}}{{d\tau }} = - \frac{9}{{2{\tau ^2}}}{\left( {\frac{1}{\tau }} \right)^{ - 3/2}}{\bf{j}} - \frac{1}{{{\tau ^2}}}{\bf{k}} \cr
& {\text{ }}\frac{{d{\bf{r}}}}{{d\tau }} = - \frac{9}{{2{\tau ^{5/2}}}}{\bf{j}} - \frac{1}{{{\tau ^2}}}{\bf{k}} \cr
& {\text{Expressing }}{\bf{r}}{\text{ in terms of }}\tau {\text{ and differentiate}} \cr
& {\bf{r}} = {\bf{i}} + 3{\left( {\frac{1}{\tau }} \right)^{3/2}}{\bf{j}} + \frac{1}{\tau }{\bf{k}} \cr
& {\text{Calculate }}\frac{{d{\bf{r}}}}{{d\tau }} \cr
& \frac{{d{\bf{r}}}}{{d\tau }} = {\text{ }}\frac{d}{{d\tau }}\left[ {{\bf{i}} + 3{{\left( {\frac{1}{\tau }} \right)}^{3/2}}{\bf{j}} + \frac{1}{\tau }{\bf{k}}} \right] \cr
& \frac{{d{\bf{r}}}}{{d\tau }} = {\text{ 0}}{\bf{i}}{\text{ + }}\frac{9}{3}{\left( {\frac{1}{\tau }} \right)^{1/2}}\left( { - \frac{1}{{{t^2}}}} \right){\bf{i}} - \frac{1}{{{\tau ^2}}}{\bf{k}} \cr
& {\text{ }}\frac{{d{\bf{r}}}}{{d\tau }} = - \frac{9}{{2{\tau ^{5/2}}}}{\bf{j}} - \frac{1}{{{\tau ^2}}}{\bf{k}} \cr} $$
