Where is the mistake in the calculation of $y'$ if $ y = \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{1/4} $? Plase take a look here.
If $ y = \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{1/4} $
\begin{eqnarray}
y'&=&  \dfrac{1}{4} \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{-3/4}  \left \{  \dfrac{2x(x^2-1) - 2x(x^2+1) }{(x^2-1)^2} \right \}\\
&=& \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{-3/4} \dfrac{-x}{(x^2-1)^2}.
\end{eqnarray}
By the other hand, we have
\begin{equation}
\log y = \dfrac{1}{4} \left \{ \log (x^2+1)  - \log (x^2-1) \right \}
\end{equation}
Then, 
\begin{eqnarray}
\dfrac{dy}{dx} &=& y  \dfrac{1}{4} \left \{ \dfrac{2x}{(x^2+1)} -\dfrac{ 2x}{(x^2-1)} \right \} \\
&=&  \dfrac{1}{4} \dfrac{x^2+1}{x^2-1} \cdot 2x  \dfrac{(x^2-1) - (x^2+1)}{(x^2+1)(x^2-1)} \\
&=&  \dfrac{x^2+1}{x^2-1} \dfrac{-x}{(x^2+1)(x^2-1)} \\
&=& \dfrac{-x}{(x^2-1)^2}.
\end{eqnarray}
But this implies,
\begin{equation}
\dfrac{-x}{(x^2-1)^2} = \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{-3/4} \dfrac{-x}{(x^2-1)^2}.
\end{equation}
Where is the mistake?

I believe you forgot a power 1/4 when substituting for $y$ (in the calculation using logarithms).
Edited to explain further:
In your calculation, you write
\begin{align} \frac{dy}{dx} &= y\frac14 \left\{ \frac{2x}{(x^2+1)} - \frac{2x}{(x^2-1)} \right\} \\
&= \frac14 \frac{x^2+1}{x^2-1} \cdot 2x\frac{(x^2-1)-(x^2+1)}{(x^2+1)(x^2-1)}.
\end{align}
However, this should be
\begin{align} \frac{dy}{dx} &= y\frac14 \left\{ \frac{2x}{(x^2+1)} - \frac{2x}{(x^2-1)} \right\} \\
&= \frac14 \color{red}{\left(\color{black}{\frac{x^2+1}{x^2-1}}\right)^{\frac14}} \cdot 2x\frac{(x^2-1)-(x^2+1)}{(x^2+1)(x^2-1)}.
\end{align}
