Compare direct and logarithmic implicit differentiation

The derivative is $\dfrac{(1-6x^2)\ln x-(x-2x^3)/x}{(\ln x)^2}=\dfrac{(1-6x^2)\ln x+2x^2-1}{(\ln x)^2}$, for $x>0$, $x\ne1$.

Write $(x-2x^3)(\ln x)^{-1}$. Its derivative is $(1-6x^2)(\ln x)^{-1}-(x-2x^3)(\ln x)^{-2}/x$. Combining terms gives the same single fraction.

The chain rule gives $-\sin x/\sqrt{1-\cos^2x}=-\sin x/\sqrt{\sin^2x}$. Since $x$ is acute, $\sin x>0$, so the denominator is $\sin x$. Thus the derivative is the constant $k=-1$.

Differentiate to obtain $(y+xy\prime)e^{xy}=4y(1+y^2)y\prime$. Rearranging gives $\dfrac{dy}{dx}=\dfrac{ye^{xy}}{4y(1+y^2)-xe^{xy}}$. On the curve, substitute $e^{xy}=(1+y^2)^2$ and cancel the strictly positive factor $1+y^2$ to obtain $\dfrac{y(1+y^2)}{4y-x(1+y^2)}$.

Both sides are positive, so take logarithms: $xy=2\ln(1+y^2)$. Then $y+xy\prime=4yy\prime/(1+y^2)$. Multiply through and collect $y\prime$ to get $[4y-x(1+y^2)]y\prime=y(1+y^2)$, giving the same simplified fraction where its denominator is nonzero.