xyy'' - x(y')^2 - 4yy' = 8xy^2\\xyy'' - x(y')^2 = 4yy' + 8xy^2 / : xy^2\\
\\
\displaystyle \frac {yy'' - (y')^2} {y^2} = \displaystyle \frac {4y'} {xy} + 8\\
\\
d \big ( \displaystyle \frac {y'} {y} \big ) = \displaystyle \frac {4y'} {xy} + 8\\