The point on the curve $3\mathrm{y}=6\mathrm{x}-5{\mathrm{x}}^3$ the normal at which it passes through the origin is

Given $3\mathrm{y}=6\mathrm{x}-5{\mathrm{x}}^3$
$\mathrm{m}={\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|}_{\mathrm{P}\left({\mathrm{x}}_1,{\mathrm{y}}_{\mathrm{i}}\right)}=2-5{\mathrm{x}}_1^2$
Normal slope $=\frac{-1}{2-5{\mathrm{x}}_1^2}$
normal equation is ${\mathrm{y}}_1=\frac{-1}{2-5{\mathrm{x}}_1^2}{\mathrm{x}}_1 .........\left(1\right)$
P is on curve $3{\mathrm{y}}_1=6{\mathrm{x}}_1-5{\mathrm{x}}_1^3 ..........\left(2\right)$

From  (1) and (2) 

 solving we get point $\left(1,\frac{1}{3}\right)$