Starting with \[ \left ( y^{\prime }\right ) ^{\frac {n}{m}}=f\left ( x\right ) g\left ( y\right ) \] Find the solution \(z\) of equation \[ z^{\frac {n}{m}}=fg \] This will obtain number of solutions. For example for \(n=3,m=1\)\begin {align*} z_{1} & =\left ( fg\right ) ^{\frac {1}{3}}\\ z_{2} & =-\frac {1}{2}\left ( fg\right ) ^{\frac {1}{3}}+\frac {1}{2}i\sqrt {3}\left ( fg\right ) ^{\frac {1}{3}}\\ z_{3} & =-\frac {1}{2}\left ( fg\right ) ^{\frac {1}{3}}-\frac {1}{2}i\sqrt {3}\left ( fg\right ) ^{\frac {1}{3}} \end {align*}
Now if we assume that \(f>0,g>0\) then we can separate the \(f,g\) giving\begin {align*} z_{1} & =f^{\frac {1}{3}}g^{\frac {1}{3}}\\ z_{2} & =-\frac {1}{2}f^{\frac {1}{3}}g^{\frac {1}{3}}+\frac {1}{2}i\sqrt {3}f^{\frac {1}{3}}g^{\frac {1}{3}}\\ z_{3} & =-\frac {1}{2}f^{\frac {1}{3}}g^{\frac {1}{3}}-\frac {1}{2}i\sqrt {3}f^{\frac {1}{3}}g^{\frac {1}{3}} \end {align*}
or\begin {align*} z_{1} & =f^{\frac {1}{3}}g^{\frac {1}{3}}\\ z_{2} & =g^{\frac {1}{3}}\left ( -\frac {1}{2}f^{\frac {1}{3}}+\frac {1}{2}i\sqrt {3}f^{\frac {1}{3}}\right ) \\ z_{3} & =g^{\frac {1}{3}}\left ( -\frac {1}{2}f^{\frac {1}{3}}-\frac {1}{2}i\sqrt {3}f^{\frac {1}{3}}\right ) \end {align*}
This means\begin {align*} y^{\prime } & =f^{\frac {1}{3}}g^{\frac {1}{3}}\\ y^{\prime } & =g^{\frac {1}{3}}\left ( -\frac {1}{2}f^{\frac {1}{3}}+\frac {1}{2}i\sqrt {3}f^{\frac {1}{3}}\right ) \\ y^{\prime } & =g^{\frac {1}{3}}\left ( -\frac {1}{2}f^{\frac {1}{3}}-\frac {1}{2}i\sqrt {3}f^{\frac {1}{3}}\right ) \end {align*}
Which gives\begin {align*} \int \frac {dy}{g\left ( y\right ) ^{\frac {1}{3}}} & =\int f\left ( x\right ) ^{\frac {1}{3}}dx+c_{1}\\ \int \frac {dy}{g\left ( y\right ) ^{\frac {1}{3}}} & =\int \left ( -\frac {1}{2}f^{\frac {1}{3}}+\frac {1}{2}i\sqrt {3}f^{\frac {1}{3}}\right ) dx+c_{1}\\ \int \frac {dy}{g\left ( y\right ) ^{\frac {1}{3}}} & =\int \left ( -\frac {1}{2}f^{\frac {1}{3}}-\frac {1}{2}i\sqrt {3}f^{\frac {1}{3}}\right ) dx+c_{1} \end {align*}
There is no need to evaluate the integrals unless needed. Without the assumption \(f,g>0\) we could not separate them. Since \(\left ( fg\right ) ^{\frac {n}{m}}=f^{\frac {n}{m}}g^{\frac {n}{m}}\) is true under this condition when \(\frac {n}{m}\) is rational number. If \(\frac {n}{m}\) is an integer, then this condition is not needed and we can always factor out \(f,g\) and separate them.
The assumption \(f,g>0\) might be too strict to use but without this assumption this method can not be used.