The point on the curve $3 y=6 x-5 x^3$ the normal at which passes through the origin, is

(1, 1/3)

Let the required point be $(x_1, y_1)$

Now,

$3 y=6 x-5 x^3$

$\Rightarrow 3 \frac{d y}{d x}=6-15 x^2 \Rightarrow \frac{d y}{d x}=2-5 x^2 \Rightarrow\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=2-5 x_1^2$

The equation of the normal at $\left(x_1, y_1\right)$ is

$y-y_1=\frac{-1}{2-5 x_1^2}\left(x-x_1\right)$

If it passes through the origin, then

$0-y_1=\frac{1}{2-5 x_1{ }^2}\left(0-x_1\right) \Rightarrow y_1=\frac{-x_1}{2-5 x_1{ }^2}$            .......(i)

Since $\left(x_1, y_1\right)$ lies on the given curve. Therefore,

$3 y_1=6 x_1-5 x_1^3$            .......(ii)

Solving (i) and (ii), we obtain $x_1=1$ and $y_1=1 / 3$

Hence, the required point is (1, 1/3).