\begin {align*} \frac {dy}{dx} & =\frac {-3yx}{3x^{2}+y^{2}}\\ y\left ( 0\right ) & =1 \end {align*}
At \(x=0,y=1\) then \(f\left ( x,y\right ) =\frac {-3y-x}{3x^{2}+y^{2}}\) is defined. And \(f_{y}\) is also defined at \(x=0,y=1\). Hence a unique solution exist.
Let \(y=ux\) or \(u=\frac {y}{x}\), hence \(\frac {dy}{dx}=x\frac {du}{dx}+u\) and the above ode becomes\begin {align*} x\frac {du}{dx}+u & =\frac {-3ux^{2}}{3x^{2}+u^{2}x^{2}}\\ x\frac {du}{dx}+u & =\frac {-3u}{3+u^{2}}\\ x\frac {du}{dx} & =\frac {-3u}{3+u^{2}}-u\\ & =\frac {-3u-u\left ( 3+u^{2}\right ) }{3+u^{2}}\\ & =\frac {-6u-u^{3}}{3+u^{2}} \end {align*}
This is separable.\[ \frac {3+u^{2}}{-6u-u^{3}}du=\frac {1}{x}dx \] Integrating\begin {align*} \int \frac {3+u^{2}}{-6u-u^{3}}du & =\int \frac {1}{x}dx\\ -\frac {1}{2}\ln u-\frac {1}{4}\ln \left ( u^{2}+6\right ) & =\ln x+c\\ -\frac {1}{2}\ln u-\frac {1}{4}\ln \left ( u^{2}+6\right ) -\ln x & =\ln x+\ln c_{1} \end {align*}
Solving for \(u\) gives\begin {align*} u_{1} & =-\sqrt {-3-\frac {1}{2}\sqrt {36+\frac {4}{x^{4}c_{1}^{4}}}}\\ u_{2} & =\sqrt {-3-\frac {1}{2}\sqrt {36+\frac {4}{x^{4}c_{1}^{4}}}}\\ u_{3} & =-\sqrt {-3+\frac {1}{2}\sqrt {36+\frac {4}{x^{4}c_{1}^{4}}}}\\ u_{4} & =\sqrt {-3+\frac {1}{2}\sqrt {36+\frac {4}{x^{4}c_{1}^{4}}}} \end {align*}
Hence\begin {align*} \frac {y_{1}}{x} & =-\sqrt {-3-\frac {1}{2}\sqrt {36+\frac {4}{x^{4}c_{1}^{4}}}}\\ \frac {y_{2}}{x} & =\sqrt {-3-\frac {1}{2}\sqrt {36+\frac {4}{x^{4}c_{1}^{4}}}}\\ \frac {y_{3}}{x} & =-\sqrt {-3+\frac {1}{2}\sqrt {36+\frac {4}{x^{4}c_{1}^{4}}}}\\ \frac {y_{4}}{x} & =\sqrt {-3+\frac {1}{2}\sqrt {36+\frac {4}{x^{4}c_{1}^{4}}}} \end {align*}
or for \(x\geq 0\)\begin {align*} y_{1} & =-\sqrt {-3x^{2}-\frac {1}{2}\sqrt {36x^{4}+\frac {4}{c_{1}^{4}}}}\\ y_{2} & =\sqrt {-3x^{2}-\frac {1}{2}\sqrt {36x^{4}+\frac {4}{c_{1}^{4}}}}\\ y_{3} & =-\sqrt {-3x^{2}+\frac {1}{2}\sqrt {36x^{4}+\frac {4}{c_{1}^{4}}}}\\ y_{4} & =\sqrt {-3x^{3}+\frac {1}{2}\sqrt {36x^{4}+\frac {4}{c_{1}^{4}}}} \end {align*}
Applying IC \(y\left ( 0\right ) =1\)\begin {align*} -1 & =\sqrt {-\sqrt {\frac {1}{c_{1}^{4}}}}\\ 1 & =\sqrt {-\sqrt {\frac {1}{c_{1}^{4}}}}\\ -1 & =\sqrt {\sqrt {\frac {1}{c_{1}^{4}}}}\\ 1 & =\sqrt {\sqrt {\frac {1}{c_{1}^{4}}}} \end {align*}
or\begin {align*} -1 & =\sqrt {\frac {1}{c_{1}^{4}}}\\ -1 & =\sqrt {\frac {1}{c_{1}^{4}}}\\ 1 & =\sqrt {\frac {1}{c_{1}^{4}}}\\ 1 & =\sqrt {\frac {1}{c_{1}^{4}}} \end {align*}
Throwing the first 2 since complex. Then \(c_{1}=1\). Hence\begin {align*} y & =\sqrt {-3x^{3}+\frac {1}{2}\sqrt {36x^{4}+4}}\\ & =\sqrt {-3x^{3}+\sqrt {9x^{4}+1}} \end {align*}