The solution of the differential equation $3xy\text{'}-3y+{(x^2-y^2)}^{1/2}=0$ satisfying the condition $y\left(1\right)=1$ is

$3xy\text{'}-3y+{(x^2-y^2)}^{1/2}=0\Rightarrow 3x\frac{dy}{dx}=3y+\sqrt{x^2-y^2}\Rightarrow \frac{dy}{dx}=\frac{3y-\sqrt{x^2-y^2}}{3x}\stackrel{ }{\to }\left(1\right)$

put $y=vx$ then $\frac{dy}{dx}==v+x\frac{dv}{dx}$

$\left(1\right)\Rightarrow v+x\frac{dv}{dx}=\frac{3vx-\sqrt{x^2-v^2{x}^2}}{3x}=\frac{3v-\sqrt{1-v^2}}{3}\Rightarrow x\frac{dv}{dx}=\frac{3v-\sqrt{1-v^2}}{3}-v=-\frac{\sqrt{1-v^2}}{3}$

$\Rightarrow \frac{-3}{\sqrt{1-v^2}}dv=\frac{dx}{x}\Rightarrow 3{\cos }^{-1}v=\log \left|x\right|+c\Rightarrow 3{\cos }^{-1}\frac{y}{x}=\log \left|x\right|+c$

Now, $y\left(1\right)=1\Rightarrow c=0$

$\therefore$The required solution is $3{\cos }^{-1}\frac{y}{x}=\log \left|x\right|$