Answer
The osculating circle can be parametrized by
${\bf{o}}\left( t \right) = \left( {\frac{1}{{\sqrt 2 }}\cos t + \frac{1}{2},\frac{1}{{\sqrt 2 }}\sin t - \frac{1}{2}} \right)$
Work Step by Step
Write $f\left( x \right) = y = {x^2} - {x^3}$. Thus, we can use the parametrization:
${\bf{r}}\left( x \right) = \left( {x,f\left( x \right)} \right) = \left( {x,{x^2} - {x^3}} \right)$
Step 1. Find the radius of the osculating circle
By Eq. (5) of Theorem 2 (Section 14.4), the curvature is
$\kappa \left( x \right) = \frac{{\left| {f{\rm{''}}\left( x \right)} \right|}}{{{{\left( {1 + f'{{\left( x \right)}^2}} \right)}^{3/2}}}} = \frac{{\left| {2 - 6x} \right|}}{{{{\left( {1 + {{\left( {2x - 3{x^2}} \right)}^2}} \right)}^{3/2}}}}$
$\kappa \left( 1 \right) = \frac{4}{{2\sqrt 2 }} = \sqrt 2 $
Thus, the osculating circle has radius $R = 1/\kappa \left( 1 \right) = 1/\sqrt 2 $.
Step 2. Find the unit normal vector ${\bf{N}}$ at $x=1$
We have ${\bf{r}}'\left( x \right) = \left( {1,2x - 3{x^2}} \right)$. The unit tangent vector is
${\bf{T}}\left( x \right) = \frac{{{\bf{r}}'\left( x \right)}}{{||{\bf{r}}'\left( x \right)||}} = \frac{{\left( {1,2x - 3{x^2}} \right)}}{{||\left( {1,2x - 3{x^2}} \right)||}}$
${\bf{T}}\left( x \right) = \frac{1}{{\sqrt {1 + 4{x^2} - 12{x^3} + 9{x^4}} }}\left( {1,2x - 3{x^2}} \right)$
${\bf{T}}\left( 1 \right) = \frac{1}{{\sqrt 2 }}\left( {1, - 1} \right)$
Since ${\bf{N}}\left( 1 \right)$ must be orthogonal to ${\bf{T}}\left( 1 \right)$, we obtain ${\bf{N}}\left( 1 \right) = \frac{1}{{\sqrt 2 }}\left( { - 1, - 1} \right)$ because their dot product is zero. Notice that the ${\bf{N}}\left( 1 \right)$ is an unit vector and points inward of the curve.
Step 3. Find the center of the osculating circle $Q$
By Eq. (9) of Section 14.4, the center of the osculating circle at $x=1$ is
$\overrightarrow {OQ} = {\bf{r}}\left( 1 \right) + R{\bf{N}}\left( 1 \right) = \left( {1,0} \right) + \frac{1}{{\sqrt 2 }}\cdot\frac{1}{{\sqrt 2 }}\left( { - 1, - 1} \right)$
$\overrightarrow {OQ} = \left( {\frac{1}{2}, - \frac{1}{2}} \right)$
Step 4. Parametrize the osculating circle
At $x=1$, the osculating circle has radius $R = \frac{1}{{\sqrt 2 }}$ and center $\left( {\frac{1}{2}, - \frac{1}{2}} \right)$. So, the osculating circle can be parametrized by
${\bf{o}}\left( t \right) = \left( {\frac{1}{2}, - \frac{1}{2}} \right) + \frac{1}{{\sqrt 2 }}\left( {\cos t,\sin t} \right)$
${\bf{o}}\left( t \right) = \left( {\frac{1}{{\sqrt 2 }}\cos t + \frac{1}{2},\frac{1}{{\sqrt 2 }}\sin t - \frac{1}{2}} \right)$
