Solution:

$F = {xy + y^2-2y^2y'}$ -----------(1)

differentiating (1) w.r.t. y

$\frac{\delta F}{\delta y} = x+2y-4yy'$--------(2)

differentiating (1) w.r.t. y'

$\frac{\delta F}{\delta y} = 0+0-2y^2$

$\frac{\delta F}{\delta y} = -2y^2$-------(3)

Using Euler's equation

$\frac{\delta F}{\delta y} -\frac {d}{dx} (\frac{\delta F}{\delta y'}) = 0 $ ------(4)

Put (2),(3) in (4),

$\begin{aligned} x+2y -4yy'-\frac {d}{dx}(-2y^2) &= 0 \\ x +2y - 4yy'-4y\frac {dy}{dx} &= 0 \\ x +2y - 4yy'-4yy' &= 0 \\ x+2y &= 0 \\ y &= \frac{-x}{2} \end{aligned}$

This is required extremal.