$\frac{d}{dx}\left(e^{2y}=\ln\left(x^3+2y\right)\right)$

$\left(\frac{1}{2}x^2+\frac{2}{3}y\right)\left(\frac{1}{2}x^2-\frac{2}{3}y\right)$

$\frac{dy}{dx}=-15x^2=8$

$\ln\left(2\cosh\left(3x\right)\right)$

$2x\frac{dy}{dx}-\left(\frac{dy}{dx}\right)^2y-y=0$

$x^2+x-5=0$