`f(x,y)=x^3+y^3=2xy`
Find Implicit differentiation, Derivative with respect to x

Solution:
Method-1: using Chain Rule
`x^3+y^3-2xy=0`

After implicit differentiation

`d/(dx)(x^3+y^3-2xy)=3x^2+3y^2y'-2y-2xy'`

`d/(dx)(x^3+y^3-2xy)`

`=d/(dx)(x^3)+d/(dx)(y^3)-d/(dx)(2xy)`

`d/(dx)(y^3)=3y^2y'`

`d/(dx)(y^3)`

`=3y^2*d/(dx)(y)`

`=3y^2*y'`

`=3y^2y'`

`d/(dx)(2xy)=2y+2xy'`

`d/(dx)(2xy)`

`=2*(d/(dx)(x))y+2x(d/(dx)(y))`

`=2*1y+2x(y')`

`=2y+2xy'`

`=3x^2+3y^2y'-(2y+2xy')`

`=3x^2+3y^2y'-2y-2xy'`

`3x^2+3y^2y'-2y-2xy'=0`

`=>3y^2y'-2xy'+3x^2-2y = 0`

`=>3y^2y'-2xy' = -3x^2+2y`

`=>y'(3y^2-2x) = -3x^2+2y`

`=>y' = (-3x^2+2y)/(3y^2-2x)`

`=>y' = ((-3x^2+2y)/(3y^2-2x))`

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Method-2: using partial derivatives `Fx,Fy`. If `F(x,y)=0`, then `y'=-(Fx)/(Fy)`

`Fx=(del)/(del x)(x^3+y^3-2xy)=3x^2-2y`

Now,
`del/(delx)(x^3+y^3-2xy)`

`=del/(delx)(x^3)+del/(delx)(y^3)-del/(delx)(2xy)`

`del/(delx)(y^3)=0`

`del/(delx)(y^3)`

`=3y^2*del/(delx)(y)`

`=3y^2*0`

`=0`

`del/(delx)(2xy)=2y+0`

`del/(delx)(2xy)`

`=2*(del/(delx)(x))y+2x(del/(delx)(y))`

`=2*1y+2x*0`

`=2y+0`

`=3x^2+0-(2y+0)`

`=3x^2+0-2y-0`

`=3x^2-2y`

`:.(del)/(del x)(x^3+y^3-2xy)=3x^2-2y`

`Fy=(del)/(del y)(x^3+y^3-2xy)=3y^2-2x`

Now,
`del/(dely)(x^3+y^3-2xy)`

`=del/(dely)(x^3)+del/(dely)(y^3)-del/(dely)(2xy)`

`del/(dely)(x^3)=0`

`del/(dely)(x^3)`

`=3x^2*del/(dely)(x)`

`=3x^2*0`

`=0`

`del/(dely)(2xy)=0+2x`

`del/(dely)(2xy)`

`=2*(del/(dely)(x))y+2x(del/(dely)(y))`

`=2*0y+2x*1`

`=0+2x`

`=0+3y^2-(0+2x)`

`=0+3y^2-0-2x`

`=3y^2-2x`

`:.(del)/(del y)(x^3+y^3-2xy)=3y^2-2x`

`y'=-(Fx)/(Fy)`

`:.y'=-(3x^2-2y)/(3y^2-2x)`

`:.y'=(-(3x^2-2y))/(3y^2-2x)`

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