Answer
$y' = \dfrac{1-2xy^2 \cos(x^2y^2)}{2x^2y\cos (x^2y^2)}$
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
$\sin(x^2y^2)-x=0$
Then derivate the whole equation. Rember the put y' every time you derivate y
$(2xy^2+2x^2yy')\cos(x^2y^2)-1=0$
*Note: In this one, you have to apply the chain rule
Solve for y' and you have the answer
$y'(2x^2y\cos (x^2y^2))=1-2xy^2 \cos(x^2y^2)$
$y' = \dfrac{1-2xy^2 \cos(x^2y^2)}{2x^2y\cos (x^2y^2)}$
