\[ y^{\prime }=-\frac {y\left ( y^{2}+3x^{2}+2x\right ) }{x^{2}+y^{2}}\]
Applying change of variables \(y=ux\) results in\[ u^{\prime }=-\frac {u\left ( u^{2}+3\right ) }{u^{2}+1}\frac {x+1}{x}\] Which is separable. Solving this for \(u\left ( x\right ) \) by integration gives\begin {align*} \int \frac {1}{-\frac {u\left ( u^{2}+3\right ) }{u^{2}+1}}du & =\int \frac {x+1}{x}dx\qquad -\frac {u\left ( u^{2}+3\right ) }{u^{2}+1}\neq 0\\ \frac {1}{3}\ln \left ( \left ( u^{2}+3\right ) u\right ) +x+\ln \left ( x\right ) & =c_{1} \end {align*}
Hence the solution in \(y\left ( x\right ) \) is\[ \frac {1}{3}\ln \left ( \left ( \left ( \frac {y}{x}\right ) ^{2}+3\right ) \frac {y}{x}\right ) +x+\ln \left ( x\right ) =c_{1}\]
Singular solution is when \(u\left ( u^{2}+3\right ) =0\) or \(u=0,u=\pm i\sqrt {3}\) which implies \(y=0\) and \(y=\pm i\sqrt {3}x\). Hence the solutions are
\begin {align*} \frac {1}{3}\ln \left ( \left ( \left ( \frac {y}{x}\right ) ^{2}+3\right ) \frac {y}{x}\right ) +x+\ln \left ( x\right ) & =c_{1}\\ y & =0\\ y & =i\sqrt {3}x\\ y & =-i\sqrt {3}x \end {align*}