Finding minima and maxima of $\frac{e^{1/({1-x^2})}}{1+x^2}$ 
Find minima and maxima of $\frac{e^{1/({1-x^2})}}{1+x^2}$.

I have:
\begin{align}
f'(x)=\frac{ 2x\cdot  e^{{1}/({1-x^2})} +\left(\frac{1+x^2}{(1-x^2)^2}-1\right)}{(1+x^2)^2}.
\end{align}
I have $x=0$ and $x=+\sqrt{3},x=-\sqrt{3}$ for solutions of $f'(x)=0$, but I can't find $f''(x)$ so I need help if someone can simplify this?
 A: Let's first start by correcting your calculation of $f'(x)$.
Recall the quotient rule. Given $$f(x) = \frac{g(x)}{h(x)}$$
and $h(x)\neq 0,$ then 
$$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$$
Now, for $$f(x) = \dfrac{\overbrace{e^{1/(1-x^2)}}^{g(x)}}{\underbrace{1+x^2}_{h(x)}}$$we have $g(x) = e^{1/(1-x^2)}$ and $h(x) = 1 + x^2$.
$g'(x) = \left(\dfrac 1{1-x^2}\right)' e^{1/(1-x^2)} = \dfrac{-2x}{(1 - x^2)^2}e^{1/(1 - x^2)}$ and $h'(x) = 2x$
A: We calculate the derivative, because that is the first thing you did. It is fairly complicated. The denominator is $(1+x^2)^2$. The numerator, after some simplification, turns out to be
$$\frac{2x^3(3-x^2)}{(1-x^2)^2}e^{-1/(1-x^2)}.$$
This is indeed $0$ at $x=0$ and $x=\pm\sqrt{3}$.
Looking at the function: Note the symmetry of $y=f(x)$ about the $y$-axis. Then take note of the singularities at $x=\pm 1$. We look only at $x=1$, since symmetry takes care of the other. If $x$ is a little under $1$, then $1-x^2$ is small positive, and therefore $e^{1/(1-x^2)}$ is huge positive.
On the other side of $1$ but near $1$, we are looking at $e$ to a huge negative power. The result is nearly $0$. Division by $(1-x^2)^2$ still leaves us nearly at $0$. 
Note also that for $x$ big, our function is very close to $0$.
These observations tell us everything: we will have a local minimum at $x=0$. The value there is $e$. We will have local maxima at $x=\pm \sqrt{3}$. Amusingly, the values there are *less than $e$.
Note that the local minimum is not a global minimum: For $x$ close to $1$ but to the right of $1$, or for $x$ close to $-1$ but to the left of $-1$, the function is close to $0$. The function is also positive and close to $0$ when $x$ is large positive or negative. So there is no global minimum.
Using the derivative: If you wish, we can obtain this information from the derivative. The denominator of the derivative is safely  positive, as is $e^{-1/(1-x^2)}$. So we need only look at the $2x^3(3-x^2)$ part. Look first near $0$.  
In the interval $(-1,0)$ the derivative is negative, and in the interval $(0,1)$ it is positive. so the function decreased, then increased, so reached a local min at $x=0$.
Now look near $x=\sqrt{3}$. In the interval $(1,\sqrt{3})$ the derivative is positive, and in $(\sqrt{3},\infty)$ is is nrgative, so we reached a local max at $x=\sqrt{3}$.
The story for $-\sqrt{3}$ is the same. 
Summary: We have a local min at $x=0$, and local maxima at $x=\pm\sqrt{3}$. There is no global min, and there is no global max. 
Remark: We did not compute the second derivative. For one thing, we are not that heroic, or masochistic. Students tend to overuse the second derivative test in one-variable calculus. The full story is already contained in the first derivative.
