A)
$\displaystyle y'siny=cosx(2cosy-sin^{2}x)$
B)
$\displaystyle x^{2}y'-1=cos2y $
how to solve these
Original DE:
$\displaystyle y'\sin(y)=\cos(x)(2\cos(y)-\sin^{2}(x)).$
Substitution: $\displaystyle u=\cos(y).$ Then $\displaystyle -u'=\sin(y)y'.$
Plugging into the DE yields
$\displaystyle -u'=\cos(x)(2u-\sin^{2}(x))\quad\to\quad -u'-2u\cos(x)=-\cos(x)\sin^{2}(x)$
$\displaystyle \to\quad u'+2u\cos(x)=\cos(x)\sin^{2}(x).$
I think you forgot to distribute the cosine everywhere. So, propagate these changes: what do you get?