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 $y'$?
 A: So you have the expression 
$$ 1 + y^{2} + 2xy(\dfrac{dy}{dx}) = \dfrac{2xy + x^{2}(\dfrac{dy}{dx})}{1 + x^{4}y^{2}}$$
And we can break up the term on the left hand side in this way: $$\dfrac{2xy + x^{2}(\dfrac{dy}{dx})}{1 + x^{4}y^{2}} = \dfrac{2xy }{1 + x^{4}y^{2}} + \dfrac{x^{2}(\dfrac{dy}{dx})}{1 + x^{4}y^{2}} $$
Then, subtracting $\dfrac{2xy }{1 + x^{4}y^{2}}$ from both sides of the original equation, and subtracting $2xy(\dfrac{dy}{dx})$ from both sides, the first equation becomes: 
$$ 1 + y^{2} - \dfrac{2xy }{1 + x^{4}y^{2}} =  \dfrac{x^{2}(\dfrac{dy}{dx})}{1 + x^{4}y^{2}} -  2xy(\dfrac{dy}{dx})$$
Finally, you can factor $\dfrac{dy}{dx}$ out from the right hand side, and divide both sides by the other factor to solve for $\dfrac{dy}{dx}$.
You should get: 
$$ \dfrac{1 + y^{2} - \dfrac{2xy }{1 + x^{4}y^{2}}}{\dfrac{x^{2}}{1 + x^{4}y^{2}} -  2xy} = \dfrac{dy}{dx} $$
A: 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 : 
$$y'=\frac{2xy-(1+y^2)(1+x^4y^2)}{2xy(1+x^4y^2)-x^2}.$$
