Answer
$y'=\dfrac{1-2y}{2x+2y-1}$
Work Step by Step
Apply rule of differentiation: $y'=a'(x)+b'(x)$
Here, $a'(x)=m'(x)n(x)+m(x)n'(x)$
Given: $2xy+y^2=x+\frac{y}{d/dx}$
$2(x'y+xy')+2yy'=1+y'$
$2(y+xy')+2yy'=1+y'$
$2y+2xy'+2yy'=1+y'$
$2xy'+2yy'-y'=1-2y$
$y'(2x+2y-1)=1-2y$
Hence, $y'=\dfrac{1-2y}{2x+2y-1}$