\[(2x-y)dx+(2y-x)dx=0\]
with the curve passing through \[(2,1)\]
cannot be solved by separation of variables, use of integrating factors, parts, substitution,...If we expand the brackets and rearrange however we get
\[2xdx+2ydy-(ydx+xdy)=0\]
We can write this as
\[d(x^2+y^2)-d(xy)=0\]
.Integration now gives
\[x^2+y^2-xy=c\]
.The curve passes through
\[(2,1)\]
so \[2^2+1^2-2 \times 2 \times 1=1=c\]
and the solution is \[x^2+y^2-xy=1\]
.