Is the function $f(x)=\frac{x^2}{1-|x|}$ convex?

Question

Is $f(x)=\frac{x^2}{1-|x|}, x\in \mathbb{R}$ convex?

Since $f''(x)=-\dfrac{2}{\left(\left|x\right|-1\right)^3}$, we can't claim $f''(x) > 0$ for all x. So $f(x)$ is not convex. But on page 56, Yurii Nesterov's Introductory Lectures on Convex Optimization: A Basic Course claimed that $f(x)$ is convex on $\mathbb{R}$. Note that $f(x)$ is not even defined for $x=1$. Did I miss something?

Answer 1

Hint/verification. (Desmos link)

Answer 2

Yes you are right indeed for $x>0$ we have

$$f(x)=g(x)=\frac{x^2}{1-x}\implies g''(x)=-\frac{2}{(x-1)^3}$$

and since $f(x)$ is even we have that it is convex for $x\in(-1,1)$.