Prove the trigonometric identity $\frac{1+\tan\left(x\right)}{1-\tan\left(x\right)}+\frac{1+\cot\left(x\right)}{1-\cot\left(x\right)}=0$

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

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

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

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

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

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

$y'=x+8$