$\int\frac{dx}{x-3y}$ when $y(x-y)^2=x$?

This is one of the reasons I love G. H. Hardy's "A Course of Pure Mathematics". The current question is problem 16, page 260. He advises to use the substitution $t=x-y$. So that $y=x/t^2$ and then $t=x-x/t^2$. Finally we get $x=t^3/(t^2-1), y=t/(t^2-1)$. Then $$dx=\frac{t^4-3t^2}{(t^2-1)^2}\,dt,\, x-3y=\frac{t^3-3t}{t^2-1}$$ so that $$\int\frac{dx}{x-3y}=\int\frac{t}{t^2-1}\,dt=\frac{1}{2}\log(t^2-1)+C=\frac{1}{2}\log\{(x-y)^2-1\}+C$$ Hardy gives further examples:

1) $y^2(x-y)=x^2$ which is rationalized by $y=tx$.

2) $(x^2+y^2)^2=a^2(x^2-y^2)$ which is rationalized by $x^2+y^2=t(x-y)$. Thus if $$(x^2+y^2)^2=2c^2(x^2-y^2)$$ then $$\int\frac{dx}{y(x^2+y^2+c^2)}=-\frac{1}{c^2}\log\left(\frac{x^2+y^2}{x-y}\right)+C$$

There is a deep theory of rational parametrization of algebraic curves given by relation $f(x,y)=0$ with $f$ being a polynomial which is at work here. Hardy just gives a glimpse of the theory of algebraic curves here.

Differentiating $$ x=y(x-y)^2\tag{1} $$ we get $$ \begin{align} 1&=y'(x-y)^2+2y(x-y)(1-y')\\ &=y'\left[(x-y)^2-2y(x-y)\right]+2y(x-y)\\ &=y'(x-3y)(x-y)+2y(x-y)\\ 1-2y(x-y)&=y'(x-3y)(x-y)\\ (x-y)^2-1&=(1-y')(x-3y)(x-y)\tag{2} \end{align} $$ Differentiating Statement-I, we get $$ \frac{(x-y)(1-y')}{(x-y)^2-1}=\frac1{x-3y}\tag{3} $$ which matches $(2)$. Therefore, with the exception of a constant of integration, Statement-I is correct. That is, dividing $(2)$ by $(x-3y)\left[(x-y)^2-1\right]$ and integrating, we get $$ \begin{align} \int\frac{\mathrm{d}x}{x-3y} &=\int\frac{(x-y)\,\mathrm{d}(x-y)}{(x-y)^2-1}\\ &=\int\frac{\frac12\,\mathrm{d}[(x-y)^2-1]}{(x-y)^2-1}\\[4pt] &=\frac12\log\left[(x-y)^2-1\right]+C\tag{4} \end{align} $$

As you surmised, Statement-II implies that $y'=0$, which is false.