Answer
$${\bf{T}}\left( e \right) = \frac{{3e}}{{\sqrt {9{e^2} + 1} }}{\bf{i}} - \frac{1}{{\sqrt {9{e^2} + 1} }}{\bf{j}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = 3t{\bf{i}} - \ln t{\bf{j}},{\text{ }}t = e \cr
& {\text{Calculate }}{\bf{r}}{\text{'}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {3t{\bf{i}} - \ln t{\bf{j}}} \right] \cr
& {\bf{r}}'\left( t \right) = 3t{\bf{i}} - \frac{1}{t}{\bf{j}} \cr
& {\text{Evaluate at }}t = e \cr
& {\bf{r}}'\left( e \right) = 3{\bf{i}} - \frac{1}{e}{\bf{j}} \cr
& {\bf{r}}'\left( e \right) = 3{\bf{i}} - \frac{1}{e}{\bf{j}} \cr
& {\text{The unit tangent vector }}{\bf{T}}\left( t \right){\text{ at }}t{\text{ is defined as}} \cr
& {\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{\left\| {{\bf{r}}{\text{'}}\left( t \right)} \right\|}} \cr
& {\text{at }}t = e \cr
& {\bf{T}}\left( e \right) = \frac{{{\bf{r}}'\left( e \right)}}{{\left\| {{\bf{r}}{\text{'}}\left( e \right)} \right\|}} \cr
& {\bf{T}}\left( e \right) = \frac{{3{\bf{i}} - \frac{1}{e}{\bf{j}}}}{{\left\| {3{\bf{i}} - \frac{1}{e}{\bf{j}}} \right\|}} = \frac{{3{\bf{i}} - \frac{1}{e}{\bf{j}}}}{{\sqrt {{{\left( 3 \right)}^2} + {{\left( { - \frac{1}{e}} \right)}^2}} }} = \frac{{3{\bf{i}} - \frac{1}{e}{\bf{j}}}}{{\sqrt {9 + \frac{1}{{{e^2}}}} }} \cr
& {\bf{T}}\left( e \right) = \frac{{3{\bf{i}} - \frac{1}{e}{\bf{j}}}}{{\frac{{\sqrt {9{e^2} + 1} }}{e}}} \cr
& {\bf{T}}\left( e \right) = \frac{{3e{\bf{i}} - {\bf{j}}}}{{\sqrt {9{e^2} + 1} }} \cr
& {\bf{T}}\left( e \right) = \frac{{3e}}{{\sqrt {9{e^2} + 1} }}{\bf{i}} - \frac{1}{{\sqrt {9{e^2} + 1} }}{\bf{j}} \cr} $$
