Please Can you solve in english

Let $u = x^2 - 1$ and $v = x^2 + 1$.
Then, find their derivatives: $u' = 2x$ and $v' = 2x$.

The derivative is $\boxed{f'(x) = \frac{4x}{(x^2 + 1)^2}}$.

This implies $4x = 0$, so $x = 0$.

The coordinate of the turning point is $\boxed{(0, -1)}$.

We use the quotient rule for $f'(x) = \frac{4x}{(x^2 + 1)^2}$.
Let $u = 4x \Rightarrow u' = 4$.
Let $v = (x^2 + 1)^2 \Rightarrow v' = 2(x^2 + 1)(2x) = 4x(x^2 + 1)$.

Factor out $4(x^2 + 1)$ from the numerator:

Since $f''(0) = 4 > 0$, the turning point $(0, -1)$ is a local minimum.
The nature of the turning point is a $\boxed{\text{local minimum}}$.