#### Answer

$y'=\dfrac{-2xy-y^2}{2xy+x^2}$

#### 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: $x^2y+xy^2=\frac{6}{d/dx}$
$(x^2)'y+x^2.y'+x'y^2+x(y^2)'=0$
$2xy+2xyy'+y^2+x^2y'=0$
$2xyy'+x^2y'=-2xy-y^2$
$y'(2xy+x^2)=-2xy-y^2$
Hence, $y'=\dfrac{-2xy-y^2}{2xy+x^2}$