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