4x^3 + ln y^2 + 2y = 2x
I don't know what those last 3 terms become after differentiating. The first is 12x^2, but what do those 3 become?
$\displaystyle \frac{d}{dx}[4x^3+\ln{y^2}+2y]=\frac{d}{dx}[2x]$
$\displaystyle 12x^2+\frac{2}{y}y'+2y'=2$
$\displaystyle 6x^2+\frac{1}{y}y'+y'=1$
$\displaystyle \frac{1}{y}y'+y'=1-6x^2$
$\displaystyle y'\left(\frac{1}{y}+1\right)=1-6x^2$
$\displaystyle y'=\frac{1-6x^2}{\frac{1}{y}+1}$
$\displaystyle y'=\frac{y-6x^2y}{1+y}$
$\displaystyle 4x^3 + \ln{y^2} + 2y = 2x$.
$\displaystyle \frac{d}{dx}(4x^3 + \ln{y^2} + 2y) = \frac{d}{dx}(2x)$
$\displaystyle 12x^2 + \frac{d}{dx}(\ln{y^2} + 2y) = 2$
$\displaystyle 12x^2 + \frac{d}{dy}(\ln{y^2} + 2y)\,\frac{dy}{dx} = 2$
$\displaystyle 12x^2 + \left(\frac{2y}{y^2} + 2\right)\,\frac{dy}{dx} = 2$
$\displaystyle \left(\frac{2}{y} + 2\right)\,\frac{dy}{dx} = 2 - 12x^2$
$\displaystyle \frac{dy}{dx} = \frac{2 - 12x^2}{\frac{2}{y} + 2}$
$\displaystyle \frac{dy}{dx} = \frac{1 - 6x^2}{\frac{1}{y} + 1}$.
You could clean this up more if you wanted.