Answer
$x=-3$ and $y=0$, resulting $\min \{R(t)\}=\frac{8}{6}=\frac{4}{3}$
Work Step by Step
The radius of curvature is the reciprocal of the curvature
\[
\frac{1}{k(x)}=R
\]
In our case the curve is given:
\[
9 y^{2}+4 x^{2}=36 \equiv \frac{y^{2}}{4}+\frac{x^{2}}{9}=1 \equiv\left(\frac{y}{2}\right)^{2}+\left(\frac{x}{3}\right)^{2}=1
\]
We use the parametrization $\frac{y}{2}=\sin t .$ and $\frac{x}{3}=\cos t$ since
\[
\frac{\left|x^{\prime} y^{\prime \prime}-x^{\prime \prime} y^{\prime}\right|}{[\sqrt{y^{\prime 2}+x^{\prime 2}}]^{3 / 2}}=k(t)
\]
where
\[
\begin{array}{ll}
-3 \sin t=x^{\prime} & 2 \cos t =y^{\prime}\\
-3 \cos t=x^{\prime \prime} & -2 \sin t=y^{\prime \prime}
\end{array}
\]
So:
\[
k(t)=\frac{\left|6 \sin ^{2} t-\left(-6 \cos ^{2} t\right)\right|}{\left[4 \cos ^{2} t+9 \sin ^{2} t\right]^{3 / 2}}=\frac{6\left|\sin ^{2} t+\cos ^{2} t\right|}{\left[4+5 \sin ^{2} t\right]^{3 / 2}}=\frac{6}{\left[4+5 \sin ^{2} t\right]}
\]
So $R(t)=\frac{1}{k(t)}=\frac{\left(5 \sin ^{2} t+4\right)^{3 / 2}}{6} .$ Since $\sin ^{2} t \geq 0, R(t)$ is minimum if
$\sin 0=t .$ (For $n \pi, n \in Z=t$)
Thus, for $0=t, 3=x$ and $0=y$ and for $\pi=t, x=-3$ and $y=0$, the min is $\min \{R(t)\}=\frac{8}{6}=\frac{4}{3}$
