Answer
$$\eqalign{
& \left( {\bf{a}} \right){\text{No maximum}} \cr
& \left( {\bf{b}} \right)\mathop {\lim }\limits_{x \to \infty } K = 0 \cr} $$
Work Step by Step
$$\eqalign{
& y = {x^{2/3}} \cr
& {\text{Calculate the curvature, use }}K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}} \cr
& {\text{Find }}y'\left( x \right){\text{ and }}y''\left( x \right),{\text{ }} \cr
& y'\left( x \right) = \frac{d}{{dx}}\left[ {{x^{2/3}}} \right] \cr
& y'\left( x \right) = \frac{2}{3}{x^{ - 1/3}} \cr
& y''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{2}{3}{x^{ - 1/3}}} \right] \cr
& y''\left( x \right) = - \frac{2}{9}{x^{ - 4/3}} \cr
& \underbrace {K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}}}_ \Downarrow \cr
& K = \frac{{\left| { - \frac{2}{9}{x^{ - 4/3}}} \right|}}{{{{\left( {1 + {{\left( {\frac{2}{3}{x^{ - 1/3}}} \right)}^2}} \right)}^{3/2}}}} \cr
& K = \frac{{\left| { - \frac{2}{9}{x^{ - 4/3}}} \right|}}{{{{\left( {1 + \frac{4}{{9{x^{2/3}}}}} \right)}^{3/2}}}} \cr
& K = \frac{{\left| { - \frac{2}{9}{x^{ - 4/3}}} \right|}}{{\frac{x}{{27}}{{\left( {9{x^{2/3}} + 4} \right)}^{3/2}}}} \cr
& K = \left| {\frac{6}{{{x^{1/3}}{{\left( {9{x^{2/3}} + 4} \right)}^{3/2}}}}} \right| \cr
& \left( {\bf{a}} \right){\text{ Find the point where }}K{\text{ is maximum}} \cr
& {\text{Find }}K'{\text{ using a CAS}} \cr
& K' = - \frac{{8\left( {9{x^{2/3}} + 1} \right)}}{{{x^{4/3}}{{\left( {1 + 9{x^4}} \right)}^{5/2}}}} \cr
& K' = 0 \cr
& 9{x^{2/3}} + 1 = 0 \cr
& {\text{No real solution, so no maximum}} \cr
& \cr
& \left( {\bf{b}} \right){\text{ Find }}\mathop {\lim }\limits_{x \to \infty } K \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{6}{{{x^{1/3}}{{\left( {9{x^{2/3}} + 4} \right)}^{3/2}}}} = 0 \cr
& \cr
& {\text{Summary}} \cr
& \left( {\bf{a}} \right){\text{No maximum}} \cr
& \left( {\bf{b}} \right)\mathop {\lim }\limits_{x \to \infty } K = 0 \cr} $$
