Solving for the zero of a multivariate 
How does one go about solving the roots for the following equation
$$x+y+z=xyz$$

There simply to many variables. Anyone have an idea ?
 A: There are infinitely many solutions. One way to show this is to solve for one of the variables. For example,
$$
\begin{align*}
& x+y+z=xyz  \\
 \Rightarrow& z-xyz=-x-y \\
\Rightarrow& z(1-xy)=-x-y \\
\Rightarrow&z=\frac{-x-y}{1-xy} \\
& \,\,\,\,=\frac{x+y}{xy-1}.
\end{align*}
$$
Choose any $x,y$ so that $xy-1 \neq 0$. For example, Let $x=2$ and $y=1$. Then $z=3$, and by inspection you will find that $x+y+z=xyz$ holds. You could write the solution up as the set of all ordered triples
$$\left\{\left( x,y, \frac{x+y}{xy-1} \right)\bigg| \, x,y \in \mathbb{R} \wedge xy \neq 1\right\}.$$
Additional Comment:
We introduced a possibility for division by zero into the solution set when we divided by $1-xy$ to get the result $z$, and this forced us to eliminate any values $x,y$ so that $xy=1$. We could have solved for one of the other two variables and encountered a similar restriction. As it turns out, it is not possible for the product of any two of $x,y,z$ to equal $1$ as was dutifully demonstrated by Will Jagy in comments.
Without loss of generality, we may assume that $xy=1$. Then 
$$x+y+z=z \Rightarrow x+y=0 \Rightarrow x=-y.$$
This is a contradiction.   
A: If we fix one of the variables, we get a hyperbola in that plane. So, for example, fixing any $z = z_0,$ this is your relationship:
$$ \left(x - \frac{1}{z_0} \right) \left(y - \frac{1}{z_0} \right) = \; 1 +  \frac{1}{z_0^2}  $$
Makes me think the surface could be connected. 
Indeed, as $|z| \rightarrow \infty,$ the curve approaches the fixed curve $xy=1.$
Meanwhile, the surface is smooth. The gradient of the function $xyz-x-y-z$ is only the zero vector at $(1,1,1)$ and $(-1,-1,-1)$  but those are not part of the surface. 
And, in the plane $x+y+z = 0,$ the surface contains the entirety of three lines that meet at $60^\circ$ angles, these being the intersections with the planes $x=0$ or $y=0$ or $z=0.$
EDIT: the surface actually has three connected components, sometimes called sheets in the context of hyperboloids. One is in the positive octant $x,y,z > 0,$ with $xy>1, xz>1, yz > 1$ in addition. The nearest point to the origin is $(\sqrt 3,\sqrt 3,\sqrt 3 ),$ and the thing gets very close to the walls as $x+y+z$ increases.
The middle sheet goes through the origin and, near there, is a monkey saddle. 
The third sheet is in the negative octant. 
About the middle sheet: you may recall that the real part of $(x+yi)^3$ is $x^3 - 3 x y^2.$ The graph $z =x^3 - 3 x y^2 $ is, near the origin, a monkey saddle, room for two legs and a tail for a monkey sitting on a horse.
If we rotate coordinates by $$ x = \frac{u}{\sqrt 3 } - \frac{v}{\sqrt 2 } - \frac{w}{\sqrt 3 }, $$
$$ y = \frac{u}{\sqrt 3 } + \frac{v}{\sqrt 2 } - \frac{w}{\sqrt 3 }, $$
$$ z = \frac{u}{\sqrt 3 } + \frac{2w}{\sqrt 3 }, $$
the surface becomes
$$ \color{magenta}{ u \; \left( 1 + \frac{v^2}{6} + \frac{w^2}{6} - \frac{u^2}{9}   \right) = \frac{1}{9 \sqrt 2} \; \left( w^3 - 3 w \, v^2 \right).}   $$
So, say within the ball $u^2 + v^2 + w^2 < 1,$ the multiplier of $u$ on the left is quite close to $1,$ and we have a monkey saddle. 
Took a bit of work to confirm: given constants $a+b+c = 0,$ and looking at 
$$ x=a+t, y = b+t, z = c+t,  $$
there are exactly three distinct real values of $t$ that solve $xyz=x+y+z.$
