Answer
$${\bf{N}}\left( 2 \right) = \frac{2}{{\sqrt {13} }}{\bf{i}} + \frac{3}{{\sqrt {13} }}{\bf{j}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = t{\bf{i}} + \frac{6}{t}{\bf{j}},{\text{ }}t = 3 \cr
& {\text{Calculate }}{\bf{r}}{\text{'}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = {\bf{i}} - \frac{6}{{{t^2}}}{\bf{j}} \cr
& {\text{Find the unit Tangent Vector }}{\bf{T}}\left( t \right) \cr
& {\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{\left\| {{\bf{r}}'\left( t \right)} \right\|}},{\text{ }}{\bf{r}}'\left( t \right) \ne 0 \cr
& {\bf{T}}\left( t \right) = \frac{{{\bf{i}} - \frac{6}{{{t^2}}}{\bf{j}}}}{{\left\| {{\bf{i}} - \frac{6}{{{t^2}}}{\bf{j}}} \right\|}} = \frac{{{\bf{i}} - \frac{6}{{{t^2}}}{\bf{j}}}}{{\sqrt {1 + \frac{{36}}{{{t^4}}}} }} \cr
& {\bf{T}}\left( t \right) = \frac{{{t^2}}}{{\sqrt {{t^4} + 36} }}{\bf{i}} - \frac{6}{{\sqrt {{t^4} + 36} }}{\bf{j}} \cr
& {\text{Find }}{\bf{T}}'\left( t \right) \cr
& {\bf{T}}'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{{{t^2}}}{{\sqrt {{t^4} + 36} }}} \right]{\bf{i}} - \frac{d}{{dt}}\left[ {\frac{6}{{\sqrt {{t^4} + 36} }}} \right]{\bf{j}} \cr
& {\text{Making the operations by hand and replacing }}\left( {{\text{or use a CAS}}} \right) \cr
& {\bf{T}}'\left( t \right) = \frac{{72t}}{{{{\left( {{t^4} + 36} \right)}^{3/2}}}}{\bf{i}} + \frac{{12{t^3}}}{{{{\left( {{t^4} + 36} \right)}^{3/2}}}}{\bf{j}} \cr
& {\text{Evaluate }}{\bf{T}}'\left( t \right){\text{ at }}t = 3 \cr
& {\bf{T}}'\left( 3 \right) = \frac{{72\left( 3 \right)}}{{{{\left( {{3^4} + 36} \right)}^{3/2}}}}{\bf{i}} + \frac{{12{{\left( 3 \right)}^3}}}{{{{\left( {{3^4} + 36} \right)}^{3/2}}}}{\bf{j}} \cr
& {\bf{T}}'\left( 3 \right) = \frac{{216}}{{{{\left( {117} \right)}^{3/2}}}}{\bf{i}} + \frac{{324}}{{{{\left( {117} \right)}^{3/2}}}}{\bf{j}} \cr
& {\bf{T}}'\left( 3 \right) = \frac{{216}}{{351\sqrt {117} }}{\bf{i}} + \frac{{324}}{{351\sqrt {117} }}{\bf{j}} \cr
& {\bf{T}}'\left( 3 \right) = \frac{8}{{13\sqrt {13} }}{\bf{i}} + \frac{{12}}{{13\sqrt {13} }}{\bf{j}} \cr
& \left\| {{\bf{T}}'\left( 3 \right)} \right\| = \sqrt {{{\left( {\frac{8}{{13\sqrt {13} }}} \right)}^2} + {{\left( {\frac{{12}}{{13\sqrt {13} }}} \right)}^2}} = \sqrt {\frac{{16}}{{169}}} = \frac{4}{{13}} \cr
& {\text{Finding the Principal Unit Normal Vector}} \cr
& {\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{\left\| {{\bf{T}}'\left( t \right)} \right\|}} \cr
& {\text{Evaluate }}{\bf{N}}'\left( t \right){\text{ at }}t = 3 \cr
& {\bf{N}}\left( 2 \right) = \frac{{{\bf{T}}'\left( 2 \right)}}{{\left\| {{\bf{T}}'\left( 2 \right)} \right\|}} = \frac{{\frac{8}{{13\sqrt {13} }}{\bf{i}} + \frac{{12}}{{13\sqrt {13} }}{\bf{j}}}}{{4/13}} \cr
& {\bf{N}}\left( 2 \right) = \frac{{13}}{4}\left( {\frac{8}{{13\sqrt {13} }}{\bf{i}} + \frac{{12}}{{13\sqrt {13} }}{\bf{j}}} \right) \cr
& {\bf{N}}\left( 2 \right) = \frac{2}{{\sqrt {13} }}{\bf{i}} + \frac{3}{{\sqrt {13} }}{\bf{j}} \cr} $$
