Hi!

$\displaystyle \text{A.}\,3xy^3-4x = 10y^2$

A little help with solving this please. I'll happily accept tips on solving implicit problems, considering I am weak at them.

Thank you!

$\displaystyle 3y^3 + x\cdot\frac{d}{dx}\left(y^3\right) + y^3\frac{d}{dx}(x) -\frac{d}{dx}(4x) = 20yy\prime$

$\displaystyle 3y^3+3xy^2y\prime + y^3 -4 = 20yy\prime$

$\displaystyle 4y^3+3xy^2y\prime - 4 = 20yy\prime$

$\displaystyle y\prime(3xy^2-20y)= 4-4y^3$

$\displaystyle y\prime = \frac{4-4y^3}{3xy^2-20y}$