Answer
$y'=\dfrac{-y^2\sin(xy^2)}{2xy\sin(xy^2)+1}$
Work Step by Step
In order to derivate this function you have to apply implicit differentation method.
First, take the function to it's f(x)=0 form
$\cos(xy^2)-y=0$
Then derivate the whole equation. Rember the put y' every time you derivate y
$-(y^2+2xyy')\sin(xy^2)-y'=0$
*Note: In this one, you have to apply the chain rule
Solve for y' and you have the answer:
$-y'(2xy\sin(xy^2)+1) = y^2\sin(xy^2)$
$y'=\dfrac{-y^2\sin(xy^2)}{2xy\sin(xy^2)+1}$
