$\displaystyle x^4+3x^2y^3-y^2=5$
Thanks!
$\displaystyle x^4+3x^2y^3-y^2=5$
$\displaystyle (4x^3)+[(3x^2)(3y^2dy/dx)+(y^3)(6x)]-2ydy/dx=0$
$\displaystyle (4x^3)+(9x^2y^2)dy/dx+(6xy^3)-(2y)dy/dx=0$
$\displaystyle (4x^3)+(6xy^3)=(2y)dy/dx-(9x^2y^2)dy/dx$
$\displaystyle (4x^3+6xy^3)/2y-9x^2y^2)=dy/dx$
Thanks for the help Pink!