Find the derivative using the quotient rule $\frac{d}{dx}\left(\frac{7}{2\ln\left(\frac{4}{3}\right)\mathrm{sech}\left(2x\right)}\right)$

$\left(1+x^4\right)\cdot dy+x\cdot\left(1+4y^2\right)\cdot dx=0$

$\frac{dy}{dx}=\frac{2x}{3y^2}$

$\frac{dx}{dy}=y+xy$

$dx+e^{3x}dy=0$

$factor\left(x^4+x^3-6x^2-4x+8\right)$

$\frac{dy}{dx}=\frac{3x^2+4x+2}{2\left(y+1\right)}$

$\frac{dy}{dx}=e^{3x+2y}$