Definition 1.2.12 of the textbook tells us that a random variable is continous if the CDF can be written as the integral,

$$F(x) = \int_{-\infty}^x f(u)du$$

for some function $f: \mathbb{R} \to \mathbb{R}$, which we call the density. To get the density, we take the derivative of $F$. Since $F(x)$ is a constant for $x < 0$, the derivative is zero in this range. For $x>0$, we compute the derivative as follows:

$$\frac{d}{dx}F(x) = \frac{d}{dx}e^{−1/x} = e^{−1/x}\frac{d}{dx}−1/x = \frac{e^{−1/x}}{x^2}$$

Proving that the derivative of $F$ exists at 0 takes a bit of extra calculus. In fact, we don't actually need to worry about this single point, because changing the value of a function $f$ at a single point does not change its integrals. So we can write down,

$$f(x) = \begin{cases} 0, & x \le 0\\ \frac{e^{−1/x}}{x^2}, &otherwise \end{cases}$$

And we know that the integral of this function equals $F$ as required in the first equation above.