Divide through by x: dx+y/x*sin^2(y/x)dx=sin^2(y/x)dy; dy/dx=1/sin^2(y/x)+y/x.
Let p=y/x, then dp/dx=1/x.dy/dx-y/x^2, so x^2dp/dx=xdy/dx-y and dy/dx=y/x+xdp/dx=p+xdp/dx.
So, p+xdp/dx=1/sin^2p+p and xdp/dx=1/sin^2p. So sin^2pdp=dx/x.
Integrating both sides: int(sinp.sinpdp)=lnx+k, where k is the constant of integration.
On the left, let u=sinp and dv=sinpdp, then du/dp=cosp and v=-cosp. We know that "d(uv)=udv+vdu", so d(-sinpcosp)=sin^2pdp-cospsinpdp and int(sin^2pdp)=-sinpcosp+int(cospsinpdp).
Let z=sinp then dz/dp=cosp so dz=cospdp so int(sinpcospdp)=int(zdz)=(1/2)z^2=(sin^2p)/2.
So int(sin^2pdp)=-sinpcosp+(sin^2p)/2=sin^2(y/x)-sin(y/x)cos(y/x), substituting for p, and this is equal to ln(x)+k.
The complete answer is ln(x)=sin^2(y/x)-(sin(y/x))(cos(y/x))+C where C=-k. This is an implicit solution because x and y cannot be separated to put one in terms of the other.
(Using trig identities we could have made substitutions for sin(y/x)cos(y/x) as (1/2)sin(2y/x), for example, but this is a calculus problem and not a trig problem.)