Is there any way to prove that the following space is a $2$-dimensional embedding manifold in $\mathbb{R}^3$? How I can prove that:
$X := \{(x, y, z) ∈ \mathbb{R}^3 \mid x^3 + y^3 + z^3 − 3xyz =1\}$
 is a embedded manifold that $2$-dimensional  in $\mathbb{R}^3$?
 A: Not an answer, but a hint
Lee Mosher has advised using the regular value theorem to prove the assertion. I've observed that what's needed is to show that for 
$$
f(x, y, z) = x^3 + y^3 + z^3 - 3xyz -1
$$
the set $M = f^{-1}(0)$ is a manifold. 
OP asks "How do we compute $f^{-1}$ in $\Bbb R^3$?" 
If OP means "what's a formula for $f^{-1}$?", the answer is that there isn't one: for a given value in $\Bbb R$ like, say, $0$, there are many points with $f(x, y, z) = 0$. For instance, $f(1,0,0) = f(0,1,0) = f(0,0,1) = 0$. So there's no formula (or even any function taking $0 \in \Bbb R$ to a single point $f^{-1}(0) \in \Bbb R^3$. 
On the other hand, there is a preimage of $0$ in $\Bbb R^3$ --- it's the set of all triples that are sent to zero. In other words, $f^{-1}(0)$ is defined to mean
$$
M = \{(a, b, c) \in \Bbb R^3 \mid f(a, b, c) = 0 \}.
$$
Maybe $OP$ was asking, "Yes, but how do I actually find that set, because to apply the theorem, I have to show that for every point $(a,b,c)$ in that set, the derivative is nonsingular, so I have to know all the points!" 
That's the clever part of this kind of question: often you don't have to be able to explicitly write down every element. To show that the derivative is nonsingular at some point $(a,b,c)$, it's often sufficient to know that $f(a,b,c) = 0$. 
Let me do an example. I'm going to prove that the unit circle in the plane is a 1-manifold, using the regular value theorem. I let
$$
g(x,y) = x^2 + y^2 - 1
$$
and I claim that $S = g^{-1}(0)$ is a 1-manifold. To show this using the theorem, I need to show that $dg(a,b)$ has rank $1$ for every point $(a,b) \in S$. 
I could do this easily by just writing down all the point (using sine and cosine), but I'm going to do it without that. Here goes. 
At any point $(x, y)$, $dg(x, y) = (2x, 2y)$. So for a point $(a, b) \in S$, we have $dg(a, b) = (2a, 2b)$. We want to show that this $1 \times 2$ matrix has rank $1$. So suppose that for some $(a,b) \in S$, we have that $(a,b)$ has rank $0$ instead of rank $1$. A rank-zero $1 \times 2$ matrix must be the zero vector, so we have
$$
(2a, 2b) = (0,0)
$$
from which we conclude that $a = b = 0$, which is a contradiction, because we know that $g(a, b) = a^2 + b^2 - 1$ is zero, but $g(0,0) = 0^2 + 0^2 - 1 = -1$. Hence for $(a,b) \in g^{-1}(0)$, we have that $dg(a,b)$ is rank 1, so $g^{-1}(0)$ is a submanifold. 
