\[ y^{\prime }=\frac {y}{x}-\frac {2}{x}\sqrt {\sin \left ( 3\frac {y}{x}\right ) }\]
The first step is to see if we can write the above as \begin {equation} y^{\prime }=\frac {y}{x}+g\left ( x\right ) f\left ( b\frac {y}{x}\right ) ^{\frac {n}{m}} \tag {1} \end {equation} Hence \begin {equation} y^{\prime }=\frac {y}{x}-\frac {2}{x}\left ( \sin \left ( 3\frac {y}{x}\right ) \right ) ^{\frac {1}{2}} \tag {2} \end {equation}
Comparing (2) to (1) shows that\begin {align*} n & =1\\ m & =2\\ g\left ( x\right ) & =-\frac {2}{x}\\ b & =3\\ f\left ( b\frac {y}{x}\right ) & =\sin \left ( 3\frac {y}{x}\right ) \end {align*}
Hence the solution is \begin {equation} y=ux \tag {A} \end {equation} Where \(u\) is the solution to \begin {equation} u^{\prime }=\frac {1}{x}g\left ( x\right ) f\left ( u\right ) ^{\frac {1}{2}} \tag {3} \end {equation} Therefore \(f\left ( u\right ) =\sin \left ( 3u\right ) \) and \(\left ( 3\right ) \) becomes\[ u^{\prime }=-\frac {2}{x^{2}}\sin \left ( 3u\right ) ^{\frac {1}{2}}\] This is separable. \begin {align*} \frac {1}{\sqrt {\sin \left ( 3u\right ) }}du & =-\frac {2}{x^{2}}dx\\ \int \frac {1}{\sqrt {\sin \left ( 3u\right ) }}du & =-2\int \frac {1}{x^{2}}dx\\ \int \frac {1}{\sqrt {\sin \left ( 3u\right ) }}du & =\frac {2}{x}+c_{1} \end {align*}
Leaving the integral as is, since it is too complicated to solve, then using \(y=ux\) where \(u\) is the solution of the above.