$\int\frac{dx}{x-3y}$ when $y(x-y)^2=x$? 
If y is a function of x such that $y(x-y)^2=x$
Statement-I: $$\int\frac{dx}{x-3y}=\frac12\log[(x-y)^2-1]$$
  Because
Statement-II: $$\int\frac{dx}{x-3y}=\log(x-3y)+c$$
Question: Is Statement-I true? Is Statement-II true? Is Statement-II a correct explanation for Statement-I?

I can say that II is false because y is not a constant but indeeed a function, I don't know answers to other two questions.I am thinking that probably we have to eleiminate y from the given functional equation or do a clever substitution?  

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.
