First, we have that $x \in \mathbb{R} \setminus (\mathbb{Z} \cup [0, 1))$. The given equation is equivalent to:
$$\frac{[x] + \{x\}}{[x]\{x\}} = -\frac{1}{x} \Leftrightarrow -x^2 = [x]\{x\}. \quad (*)$$
Since $-x^2 \le 0$ and $x \ne 0$, we have $[x]\{x\} < 0$, hence $[x] \le -1$. Because $[x] = -k$, with $k \in \mathbb{N}^*$, equation $(*)$ becomes $-x^2 = -k(k+x)$, which is equivalent to $x^2 - kx - k^2 = 0$, having solutions $x_{1,2} = \frac{k \pm k\sqrt{5}}{2}$.
From $[x] = -k$, we obtain that the only possible solution is $x_1 = \frac{k-k\sqrt{5}}{2}$.
Finally, we must find the values of $k \in \mathbb{N}^*$ for which $\left[\frac{k-k\sqrt{5}}{2}\right] = -k$, which is equivalent to: $-k \le k \frac{1-\sqrt{5}}{2} < -k+1 \Leftrightarrow 3k \ge k\sqrt{5} > 3k-2$, from which we obtain $k < \frac{3+\sqrt{5}}{2}$, i.e., $k \in \{1, 2\}$. From here we obtain the solutions $x_1^* = \frac{1-\sqrt{5}}{2}$ and $x_2^* = 1-\sqrt{5}$, which satisfy the given equation.