I've been trying to figure it out, read quite a bit on it, but still struggling a little bit.

$\displaystyle 2x^6+y^3-2x+2y=1062933$

and

$\displaystyle 3125=(2y^4-4x^5)^5$

Here is what I worked out.

$\displaystyle 2x^6+y^3-2x+2y=1062933$

$\displaystyle 12x^5+(3y^2)y'-2+2y'=0$

$\displaystyle y'=\frac{-12x^5+2}{(3y^2)+2}$

I'm not too sure where to begin with the 2nd one. Also I'm pretty sure y' = dy/dx right?