I read this as y'tan(x)sin(2y)=sin^2(x)+cos^2(y) where y'=dy/dx.
2y'tan(x)sin(y)cos(y)=sin^2(x)+cos^2(y)
Let u=cos^2(y), then u'=-2sin(y)cos(y)y' and -u'tan(x)=sin^2(x)+u; u'tan(x)+u=-sin^2(x)
Divide through by tan(x): u'+ucot(x)=-sin(x)cos(x).
Multiply through by unknown function v(x), yet to be determined:
vu'+uvcot(x)=-vsin(x)cos(x).
Take the left-hand side and apply the product rule for differentiation: d(uv)/dx=udv/dx+vdu/dx=uv'+vu', implying that uv'=uvcot(x), i.e., v'=vcot(x). We can separate the variables: v'/v=cot(x)=cos(x)/sin(x). Let q=sin(x) then q'=dq/dx=cos(x), dq=cos(x)dx, which gives us on integration ln(v)=∫cos(x)dx/sin(x)=∫dq/q=ln(q)=ln(sin(x)), so v=sin(x). [Note: dv/dx=v'=cos(x); v'/sin(x)=cos(x)/sin(x)=cot(x); v'/v=cot(x).] The left-hand side is therefore d(usin(x))/dx. The right-hand side becomes -sin^2(x)cos(x).
So usin(x)=∫-sin^2(x)cos(x)dx. Let p=sin(x), then dp=cos(x)dx and the integral becomes -∫p^2dp=-p^3/3+K, where K is the constant of integration. The result is usin(x)=-p^3/3+K. Now we can substitute backwards: cos^2(y)sin(x)=-sin^3(x)/3+K or 3cos^2(y)sin(x)=k-sin^3(x) where k=3K, a constant.
CHECK
Differentiate: -6cos(y)sin(y)sin(x)y'+3cos^2(y)cos(x)=-3sin^2(x)cos(x);
2cos(y)sin(y)sin(x)y'-cos^2(y)cos(x)=sin^2(x)cos(x) (dividing through by -3);
2cos(y)sin(y)sin(x)y'=sin^2(x)cos(x)+cos^2(y)cos(x)=cos(x)(sin^2(x)+cos^2(y));
Dividing through by cos(x): y'tan(x)sin(2y)=sin^2(x)+cos^2(y). OK!