Is it possible to have three real numbers that have both their sum and product equal to $1$?

There are infinite solutions. $z=1-x-y$ is a plane's equation and $z=\dfrac{1}{xy}$ is a complicated curve, but their intersection presents infinitely many real points.

At first I would try some trivial values like $x=0$ or $x=1$ or $x=-1$ and check which of them works and which doesn't.

For instance, if $x=0$, then $xyz=0$ doesn't work.

If $x=1$, $x+y+z=1\rightarrow y+z=0\rightarrow z=-y$ means that $xyz=1\rightarrow y^{2}=-1$ is a complex number also doesn't work.

Then $x=-1$ is a good attempt because $y+z=2$ and $yz=-1$ looks promising. \begin{equation*} yz=-1 \end{equation*} \begin{equation*} z=-\frac{1}{y} \end{equation*} then, \begin{equation*} y+z=2 \end{equation*} \begin{equation*} y-\frac{1}{y}=2 \end{equation*} \begin{equation*} y^{2}-1=2y \end{equation*} \begin{equation*} y^{2}-2y-1=0 \end{equation*} \begin{equation*} \Delta=4+4=8 \end{equation*} \begin{equation*} y=\frac{2\pm2\sqrt{2}}{2}=1\pm\sqrt{2} \end{equation*} \begin{equation*} \therefore z=1\mp\sqrt{2} \end{equation*}

Thus, both $\left(-1,1-\sqrt{2},1+\sqrt{2}\right)$ and $\left(-1,1+\sqrt{2},1-\sqrt{2}\right)$ are valid answers.

I could not comment above for the lack of space. Here are some notes about their signals:

As noticed before none of the variables can be zero, because $xyz\neq0$.

Also the three variables cannot be all positive, because if $x>0$, $y>0$, $z>0$, then it would put them under the interval $0<x,y,z<1$, if one of them would be greater than $1$, for example, $x>1$ then the sum $x+y+z$ would be greater than $1$, since $x>1$, $y>0$, $ z>0$. The problem of that interval is $0<y<1$ would mean that $\dfrac{1}{y}>1$ and for the same reason $0<z<1$ would mean $\dfrac{1}{z}>1$. But $xyz=1$ would make $x=\dfrac{1}{y}\cdot\dfrac{1}{z}>1$ and that would be impossible, so the three variables cannot be all positive and they cannot be localized on that interval.

On the other hand, you cannot have just one negative variable or three negative variables because their product would be negative, so you must have exactly two negative and one positive variables.

$$x+y+z=1 \quad \text{and} \quad xyz=1$$

\begin{align} xy(1-x-y) &= 1 \\ xy - x^2y - xy^2 = 1 \\ xy^2 + x^2y - xy + 1 &= 0 \\ xy^2 + (x^2 - x)y + 1 &= 0 \\ y^2 + (x-1)y + \dfrac 1x &= 0 \qquad\text{{$x\ne 0$ since $xyz=1$.}}\\ y &= \dfrac 12(1-x) \pm \dfrac 12\sqrt{(x-1)^2-\dfrac 4x} \end{align}

So we need to find non negative real numbers, $n$, for which we can solve $$(x-1)^2-\dfrac 4x = n^2$$

Clearly then, there exists an $n$ for all negative values of x.

When x is positive, Wolfram alpha tells us that the single real-valued solution to $(x-1)^2-\dfrac 4x = 0$ is $x=\xi$, where $\xi =\dfrac 13 \left(2 + \sqrt[3]{53 + 6 \sqrt{78}} + \sqrt[3]{53 - 6 \sqrt{78}}\right) \approx 2.314596212276752$

A value of $n$ will exists for all $x \ge \xi$.

Working it out, the solution set is

$$\{x,y,z\} = \left\{x,\; \dfrac 12(1-x) + \dfrac 12\sqrt{(x-1)^2-\dfrac 4x},\; \dfrac 12(1-x) - \dfrac 12\sqrt{(x-1)^2-\dfrac 4x} \;\right\}$$

for all $x \in (-\infty, 0) \cup [\xi, \infty)$ where $\xi =\dfrac 13 \left(2 + \sqrt[3]{53 + 6 \sqrt{78}} + \sqrt[3]{53 - 6 \sqrt{78}}\right) \approx 2.314596212276752$

Those numbers are the intersection of the surface $x y z=1$ and the plane $x+y+z=1$. A quick plot of these surfaces shows that these surfaces do intersect and that the solution consist of a curve (so infinitely many solutions) that looks something like a three-way hyperbola.