The solution of  the D.E $3xy\text{'}-3y+{(x^2-y^2)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=0$ satisfying  the condition $y\left(1\right)=1$  is 

$\begin{array}{l}D.E:\\ 3x.\frac{dy}{dx}-3y+{(x^2-y^2)}^{\frac{1}{2}}=0\\ 3x.dy-3ydx+{(x^2-y^2)}^{\frac{1}{2}}dx=0\\ puty=vx\\ dy=vdx+xdv\\ 3x(vdx+xdv)-3vxdx+{(x^2-v^2{x}^2)}^{\frac{1}{2}}dx=0\\ 3x^2.dv+x{(1-v^2)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}dx=0\\ \frac{-3dv}{{(1-v^2)}^{\frac{1}{2}}}=\frac{dx}{x}\\ 3\int \frac{-1}{\sqrt{1-v^2}}dv=\int \frac{1}{x}dx\\ 3{\cos }^{-1}\left(v\right)=\ln \left|x\right|+c\\ 3{\cos }^{-1}\left(\frac{y}{x}\right)=\ln \left|x\right|+c\\ Giveny\left(1\right)=1\\ 3{\cos }^{-1}\left(1\right)=\ln \left(1\right)+c\\ c=0\\ -3{\cos }^{-1}\frac{y}{x}=\ln \left|x\right|\end{array}$