Single-fraction derivatives and an implicit slope

Write $y=\ln x-\dfrac12\ln(1-2x)$. Then $\dfrac{dy}{dx}=\dfrac1x-\dfrac12\cdot\dfrac{-2}{1-2x}=\dfrac1x+\dfrac1{1-2x}=\dfrac{1-x}{x(1-2x)}$. The original logarithm requires $0<x<1/2$.

Write $y=[\cos^{-1}(x^2)]^{-1}$. The chain rule gives $\dfrac{dy}{dx}=(-1)[\cos^{-1}(x^2)]^{-2}\cdot\left(-\dfrac1{\sqrt{1-(x^2)^2}}\right)(2x)=\dfrac{2x}{[\cos^{-1}(x^2)]^2\sqrt{1-x^4}}$, for $|x|<1$.

Differentiate the original relation: $\left(y^2+2xy\dfrac{dy}{dx}\right)e^{xy^2}=(x^2+2e^x)\dfrac{dy}{dx}+y(2x+2e^x)$. At $x=0$, $y=1/2$, this becomes $(1/4+0)e^0=2e^0\dfrac{dy}{dx}+\dfrac12(2e^0)$. Thus $1/4=2\dfrac{dy}{dx}+1$, giving $\dfrac{dy}{dx}=-3/8$.

Take logarithms: $xy^2=\ln y+\ln(x^2+2e^x)$. Differentiate: $y^2+2xy\dfrac{dy}{dx}=\dfrac1y\dfrac{dy}{dx}+\dfrac{2x+2e^x}{x^2+2e^x}$. At $x=0$, $y=1/2$, this again gives $1/4=2\dfrac{dy}{dx}+1$, hence $-3/8$. The logarithmic route is valid locally at the specified point, where both logged factors are positive.