We have $f = 2xy - y"^2$

$\frac{\partial f}{\partial y} = 2x, \frac{\partial f}{\partial y'} = 0, \frac{\partial f}{\partial y"} = -2y"$

Hence the equation

$\frac{\partial f}{\partial y} - \frac{\partial}{\partial x} \frac{(\partial f)}{\partial y'} + \frac{\partial^2}{\partial x^2} \bigg(\frac{\partial f}{\partial y"}\bigg) = 0 $ becomes

$2x - 0 + \frac{\partial^2}{\partial x^2} (-2y") = 0 \\ 2x - 2\frac{\partial^2}{\partial x^2}\bigg(\frac{\partial^2 y}{\partial x^2} \bigg) = 0$

This is a linear differential equation of 4th order

It's A.E. is $D^4 = 0, D = 0,0,0,0$

The C.F. is $y = c_{1}x + c_{2}x + c_{3}x^2 + c_{4}x^3$

And P.I. is $y = \frac{1}{D^4}x \\ y = \frac{1}{D^3} \int xdx = \frac{1}{D^3} \frac{x^2}{2} \\ = \frac{1}{D^2} \int \frac{x^2}{2} = \frac{1}{D^2} \frac{x^3}{6} \\ = \frac{1}{D} \int \frac{x^3}{6} dx = \frac{1}{D} \frac{x^4}{24} \\ = \int \frac{x^4}{24} dx = \frac{x^5}{5!} = \frac{x^5}{120}$

Hence, solution is $y = c_{1} + c_{2}x + c_{3}x^2 + c_{4}x^3 + \frac{x^5}{5!}$