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 ?

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 $xyz=x+y+z.$
