Answer
$y'=-\dfrac{4x+y}{x-2y}$
Work Step by Step
$2x^{2}+xy-y^{2}=2$
Differentiate each term:
$(2x^{2})'+(xy)'-(y^{2})'=(2)'$
Use the product rule to find $(xy)'$:
$4x+(x)(y)'+(y)(x)'-(2y)(y')=0$
$4x+x(y')+y-(2y)(y')=0$
Solve for $y'$:
$xy'-2yy'=-4x-y$
$y'(x-2y)=-4x-y$
$y'=\dfrac{-4x-y}{x-2y}=-\dfrac{4x+y}{x-2y}$
