\[ y-2x^{3}\tan \left ( \frac {y}{x}\right ) -y^{\prime }x=0 \]
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 {align} y-2x^{3}\tan \left ( \frac {y}{x}\right ) -y^{\prime }x & =0\nonumber \\ y^{\prime }x & =y-2x^{3}\tan \left ( \frac {y}{x}\right ) \nonumber \\ y^{\prime } & =\frac {y}{x}-2x^{2}\tan \left ( \frac {y}{x}\right ) \tag {2} \end {align}
Comparing (2) to (1) shows that\begin {align*} n & =1\\ m & =1\\ g\left ( x\right ) & =-2x^{2}\\ b & =1\\ f\left ( b\frac {y}{x}\right ) & =\tan \left ( \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 ) \tag {3} \end {equation} Therefore \(f\left ( u\right ) =\tan u\) and \(\left ( 3\right ) \) becomes\[ u^{\prime }=-2x\tan u \] This is separable. \begin {align*} \frac {1}{\tan }du & =-2xdx\\ \int \frac {1}{\tan }du & =-2\int xdx\\ \ln \left ( \sin u\right ) & =-x^{2}+c_{1}\\ \sin u & =c_{2}e^{-x^{2}}\\ u & =\arcsin \left ( c_{2}e^{-x^{2}}\right ) \end {align*}
Hence (A) becomes\[ y=x\arcsin \left ( c_{2}e^{-x^{2}}\right ) \]