Answer
$$\left( {0,1} \right)$$
Work Step by Step
$$\eqalign{
& y = 1 - {x^3} \cr
& {\text{Calculate the curvature, use }}K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}} \cr
& {\text{Find }}y'\left( x \right){\text{ and }}y''\left( x \right),{\text{ }} \cr
& y'\left( x \right) = \frac{d}{{dx}}\left[ {1 - {x^3}} \right] \cr
& y'\left( x \right) = 3{x^2} \cr
& y''\left( x \right) = \frac{d}{{dx}}\left[ {3{x^2}} \right] \cr
& y''\left( x \right) = 6x \cr
& \underbrace {K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}}}_ \Downarrow \cr
& K = \frac{{\left| {6x} \right|}}{{{{\left( {1 + {{\left[ {3{x^2}} \right]}^2}} \right)}^{3/2}}}} \cr
& {\text{Find the point where the curvature is zero}}{\text{.}} \cr
& 0 = \frac{{\left| {6x} \right|}}{{{{\left( {1 + {{\left[ {3{x^2}} \right]}^2}} \right)}^{3/2}}}} \cr
& x = 0 \cr
& y\left( 0 \right) = 1 - {\left( 0 \right)^3} \cr
& y\left( 0 \right) = 1 \cr
& {\text{The curvature is zero at the point }}\left( {0,1} \right) \cr} $$
