The normal to the curve $3y=6x-5x^3$ at the point $\left(1,\frac{1}{3}\right)$ passes through the point:

the curve is $3y=6x-5x^3$
$\begin{array}{r}\Rightarrow \frac{dy}{dx}=2-5x^2\\ {\left.\Rightarrow \frac{dy}{dx}\right|}_{\left.1,\frac{1}{3}\right)}=-3\end{array}$
So, the slope of normal will be $\frac{1}{3}$
Equation of normal at $\left(1,\frac{1}{3}\right)$ is given by
$\begin{array}{r}y-\frac{1}{3}=\frac{1}{3}(x-1)\\ \Rightarrow x-3y=0\end{array}$
The point (3,1) satisfies this equation. 
Hence, the normal passes through (3,1).