Answer
$y'=\frac{1+x^4y^2+y^2+x^4y^4-2xy}{x^2-2xy-2x^5y^3}$
Work Step by Step
$tan^{-1}(x^2y)=x+xy^2\\
\frac{x^2y'+2xy}{1+(x^2y)^2}=1+2xyy'+y^2\\
x^2y'+2xy=1+x^4y^2+2xyy'+2x^5y^3y'+y^2+x^4y^4\\
x^2y'-2xyy'-2x^5y^3y'=1+x^4y^2+y^2+x^4y^4-2xy\\
y'(x^2-2xy-2x^5y^3)=1+x^4y^2+y^2+x^4y^4-2xy\\
y'=\frac{1+x^4y^2+y^2+x^4y^4-2xy}{x^2-2xy-2x^5y^3}$
