Answer
$y'=\dfrac{-2x^3+3x^2y-xy^2+x}{y-x^3+x^2y}$
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: $x^2(x-y)^2=x^2-\frac{y^2}{d/dx}$
$(x^2)'(x-y)^2+x^2((x-y)^2)'=2x-2yy'$
$2x(x-y)^2+2x^2(x-y)(x-y)'=2x-2yy'$
This implies,
$2yy'-2x^2(x-y)y'=2x-2x(x-y)^2-2x^2(x-y)$
$y'(2y-2x^2(x-y))=2x-2x(x-y)^2-2x^2(x-y)$
$y'=\dfrac{2x-2x(x-y)^2-2x^2(x-y}{2y-2x^2(x-y)}$
Hence, $y'=\dfrac{-2x^3+3x^2y-xy^2+x}{y-x^3+x^2y}$
