The point on the curve

Given curve $9 y^{2}=x^{3} \ldots$ (1)

Differentiate w.r.t. $x$,

$18 y \frac{d y}{d x}=3 x^{2}$

$\Rightarrow \frac{d y}{d x}=\frac{x^{2}}{6 y}$

Equation of normal:

$\left(\mathrm{y}-\mathrm{y}_{1}\right)=\frac{-1}{\text { Slope of tangent }}\left(\mathrm{x}-\mathrm{x}_{1}\right)$

$\because$ it makes equal intercepts with the axes

$\therefore$ slope of the normal $=\pm 1$

$\Rightarrow x^{2}=\pm 6 y$

Squaring both the sides,

$x^{4}=\pm 36 y^{2}$

From (1),

$x=0,4$

and $y=0, \pm \frac{8}{3}$

But the line making equal intercept cannot pass through origin.

So, the required points are $\left(4, \pm \frac{8}{3}\right)$.