$x = y^3$
$y = x^{1/3}$
$y' = \frac{1}{3}x^{-2/3}$
$y'' = -\frac{2}{9}x^{-5/3}$
At (1, 1)
$y' = \frac{1}{3}(1^{-2/3}) = \frac{1}{3}$
$y'' = -\frac{2}{9}(1^{-5/3}) = -\frac{2}{9}$
Radius of Curvature formula
$\rho = \dfrac{\left[ 1 + (y')^2 \right]^{3/2}}{\left\lvert y'' \right\lvert}$
$\rho = \dfrac{\left[ 1 + \left( \frac{1}{3} \right)^2 \right]^{3/2}}{\left\lvert -\frac{2}{9} \right\lvert}$
$\rho = 5.27 ~ \text{units}$ ← answer
