(2x^2-y^2)dx=xydy; dy/dx=(2x^2-y^2)/xy=2x/y-y/x.
Let u=y/x=yx^-1, dy/dx=2/u-u, du/dx=(1/x)dy/dx-y/x^2, so xdu/dx=dy/dx-y/x, and dy/dx=xdu/dx+y/x=xdu/dx+u=2/u-u.
xdu/dx=2/u-2u=2(1/u-u)=2(1-u^2)/u.
integral(udu/(1-u^2))=integral(dx/x).
Therefore, -(1/2)ln|1-u^2|=ln|ax|, where a is a constant. This can be written: ln|ax|+(1/2)ln|1-u^2|=(1/2)ln|a^2x^2(1-u^2)|=0, thus ln|a^2x^2(1-u^2)|=0.
So, raising e to each side: a^2x^2(1-u^2)=1, 1-u^2=1/a^2x^2, u^2=1-1/a^2x^2; replacing u by y/x:
y^2=x^2-1/a^2 and y=sqrt(x^2-A^2). (Constant A replaces 1/a.)