Solving for y' in a fraction Given the equation $x+xy^2 = \tan^{-1}(x^2y)$ find $y'$.
I have tried doing this but solving for $y'$ I need some help and would like your advice.
Work so far...
$$1+y^2+2xy\left(\frac{dy}{dx}\right)= \frac{2xy+x^2\left(\frac{dy}{dx}\right)}{1+x^4y^2}$$
What can be done now to solve for $y'$?

Multiply the both sides by $1+x^4y^2$ to get
$$(1+y^2+2xyy')(1+x^4y^2)=2xy+x^2y'.$$
Now you'll get
$$(2xy(1+x^4y^2)-x^2)y'=2xy-(1+y^2)(1+x^4y^2).$$
So, simplify the following : 
$$y'=\frac{2xy-(1+y^2)(1+x^4y^2)}{2xy(1+x^4y^2)-x^2}.$$
