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break so $\displaystyle sin(y)=yx^2-x$
then for $\displaystyle y'$
$\displaystyle cos(y)(dy/dx)=(x^2(dy/dx)+2xy)-1$
$\displaystyle cos(y)(dy/dx)-x^2(dy/dx)=2xy-1$
$\displaystyle y' = (2xy-1)/cos(y)-x^2$
uhh doesn't look right... but is it right? lol