If [tex]-2x^2y+y^2=1[/tex], taking the derivative of both sides wrt [tex]x[/tex] gives
[tex]-4xy-2x^2\dfrac{\mathrm dy}{\mathrm dx}+2y\dfrac{\mathrm dy}{\mathrm dx}=0[/tex]
Solving for [tex]\dfrac{\mathrm dy}{\mathrm dx}[/tex] gives
[tex]-4xy+(-2x^2+2y)\dfrac{\mathrm dy}{\mathrm dx}=0[/tex]
[tex](-2x^2+2y)\dfrac{\mathrm dy}{\mathrm dx}=4xy[/tex]
[tex](-2x^2+2y)\mathrm dy=4xy\,\mathrm dx[/tex]
[tex](-x^2+y)\mathrm dy=2xy\,\mathrm dx[/tex]
[tex]0=2xy\,\mathrm dx+(x^2-y)\mathrm dy[/tex]
