Example 3 \begin {align} x\left ( x-y^{3}\right ) \frac {dy}{dx} & =\left ( 3x+y^{3}\right ) y\nonumber \\ \frac {dy}{dx} & =\frac {\left ( 3x+y^{3}\right ) y}{x\left ( x-y^{3}\right ) } \tag {1} \end {align} We start by checking if it is isobaric or not. Using\begin {align*} m & =\frac {f+xf_{x}}{f-yf_{y}}\\ & =\frac {\frac {\left ( 3x+y^{3}\right ) y}{x\left ( x-y^{3}\right ) }+x\left ( \frac {3y}{x\left ( -y^{3}+x\right ) }-\frac {\left ( y^{3}+3x\right ) y}{x^{2}\left ( -y^{3}+x\right ) }-\frac {\left ( y^{3}+3x\right ) y}{x\left ( -y^{3}+x\right ) ^{2}}\right ) }{\frac {\left ( 3x+y^{3}\right ) y}{x\left ( x-y^{3}\right ) }-y\left ( \frac {3y^{3}}{x\left ( -y^{3}+x\right ) }+\frac {y^{3}+3x}{x\left ( -y^{3}+x\right ) }+\frac {3\left ( y^{3}+3x\right ) y^{3}}{x\left ( -y^{3}+x\right ) ^{2}}\right ) }\\ & =\frac {-4\frac {y^{4}}{\left ( x-y^{3}\right ) ^{2}}}{-12\frac {y^{4}}{\left ( x-y^{3}\right ) ^{2}}}\\ & =\frac {1}{3} \end {align*}

\(m=\frac {1}{3}\) makes each term the same weight \(\frac {4}{3}\). Hence the substitution \(y=vx^{\frac {1}{3}}\) will make the ode separable. Substituting this in (1) results in \[ \frac {dv}{dx}=\frac {-4}{3x}\frac {v\left ( v^{3}+2\right ) }{\left ( v^{3}-1\right ) }\] Which is separable. This is solved for \(v\), and then \(y\) is found from \(y=vx^{\frac {1}{3}}\).