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