Solution:

Curve: $9 \mathrm{ay}^{2}=x(x-3 a)^{2}$-----------------(1)

The given curve is strophoid.

$\begin{aligned} \text { Differentiate eqn }(1) \text { w.r.t } x \\ 18 a y \frac{d y}{d x}=2 x(x-3 a)+(x-3 a)^{2} \end{aligned}$

$\therefore 18 a y \frac{d y}{d x}=3(x-3 a)(x-a)$

$\therefore \frac{d y}{d x}=\frac{(x-3 a)(x-a)}{6 a y}$

$\begin{aligned} \text { Squaring both the sides, } \\ \left(\frac{d y}{d x}\right)^{2}=\frac{(x-3 a)^{2}(x-a)^{2}}{36 a^{2} y^{2}} \\ \therefore\left(\frac{d y}{d x}\right)^{2} =\frac{(x-3 a)^{2}(x-a)^{2}}{4 a x(x-3 a)^{2}} \end{aligned}$

The perimeter of given curve is

$\begin{aligned} \mathrm{S} &=\int_{0}^{3 a} \sqrt{1+\left(\frac{d y}{d x}\right)^{2} d x}=\int_{0}^{3 a} \sqrt{1+\frac{(x-a)^{2}}{4 a x}} d x=\int_{0}^{3 a} \sqrt{\frac{(x+a)^{2}}{4 a x}} d x \\ \therefore \mathrm{S} &=\int_{0}^{3 a} \frac{x+a}{2 \sqrt{x} \sqrt{a}} d x \\ \therefore \mathrm{S} &=\frac{1}{2 \sqrt{a}} \int_{0}^{3 a}\frac{ x+a}{\sqrt{x}} d x \end{aligned}$

$=\frac{1}{2 \sqrt{a}}\left[\frac{2 x \sqrt{x}}{3}+2 \sqrt{x}\right]_{0}^{3 a}$

$=\frac{1}{2 \sqrt{a}}\left(\frac{2 a \sqrt{3 a}}{1}+2 \sqrt{3 a}\right)$

$\therefore S=2 \sqrt{3}$---------------(Half Curve length)

$\therefore$ The totallength of given curve $=2 \mathrm{S}=4 \sqrt{3}$ units.