\[\begin{array}{*{35}{l}}
\Rightarrow ~x\text{ }{{\left( y ‘\right)}^{2}}~+\text{ }xyy’’\text{ }-\text{ }yy’\text{ }=\text{ }0 \\
\Rightarrow ~xyy’’\text{ }+\text{ }x{{\left( y’ \right)}^{2}}~-\text{ }yy’\text{ }=\text{ }0 \\
\end{array}\]
Hence, xyy” + x (y’)2 – yy’ = 0is the required differential equation.