The universal enveloping algebra of the Heisenberg Lie algebra Let $K$ be an algebraically closed field of characteristic zero. Let $U=K \langle x_1, x_2, x_3 \  | \ \ [x_1,x_2]=x_3, [x_1,x_3]=0, [x_2,x_3]=0 \rangle$   be the enveloping  algebra of the heisenberg Lie algebra. Does this rule
$$ [f,g]= \sum_{i=1}^3\sum_{j=1}^3 [x_i,x_j]\frac{\partial f}{\partial x_i}\frac{\partial g}{\partial x_j}$$
hold in U?
where $\frac{\partial f}{\partial x_i}$ is the standard derivetive of $f$ with respect to $x_i$.
I know it  holds in the symmetric algebra of any finit dimensional Lie algebra. But can we say since it holds in the symmetric algebra then it holds in the enveloping algebra. Then we can use it for computing $[f,g]$ in the enveloping algebra.
For example in case this rule holds, then I can use it for example to answer this question see here which asked about  the center of our enveloping algebra $U$. In case this rule above is correct, our answer for their question would be as follow:
An element $f$ is in the center of the algebra $U$ if and only if
$$ 0=[x_1, f]=x_3 \frac{\partial f}{\partial x_2}  \ \ {\rm and} \ \  0=[x_2, f]=-x_3 \frac{\partial f}{\partial x_1} \ \ \ \ \ \ \ \ \ ({\rm by \ \ using \ the \  rule \ above})$$ if and only if $\frac{\partial f}{\partial x_1}=0$ and $\frac{\partial f}{\partial x_2}=0$ if and only if $f \in K[x_3]$. Therefore, the center of U is $K[x_3]$.
 A: Question: "But can we say since it holds in the symmetric algebra then it holds in the enveloping algebra. Then we can use it for computing [f,g] in the enveloping algebra."
Suggestion: You should look up the notion "Poisson algebra" - maybe it is related to your construction - your product seems similar to a "Poisson product".
A Poisson algebra is a commutative unital k-algebra B with a k-Lie bracket  $\{, \}$
such that for all $a,b,c\in B$ it follows
$$\{ab,c\}=a\{b,c\}+b\{a,c\}.$$
Hence for any $c \in B$ you get an operator $\alpha(x):=\{x,-\}\in Der_k(B)$.
This gives a map of $k$-Lie algebras
$$\alpha: B \rightarrow Der_k(B).$$
The Poisson product $\{,\}$ has the property that
$$\{x,by\}=b\{x,y\}+\alpha(x)(b)y.$$
Hence $\{B, \alpha\}$ is "similar" to a Lie algebroid: The map is $k$-linear but not $B$-linear.
If $A, A_i$ is an $\mathbb{N}$-filtered almost commutative associative $k$-algebra, it follows $gr(A,A_i)$ is a Poisson algebra.
Example:  If $\mathfrak{g}$ is a $k$-Lie algebra it follows $U(\mathfrak{g})$ is an $\mathbb{N}$-filtered almost commutative associative $k$-algebra, hence $Gr(U(\mathfrak{g})) \cong Sym^*_k(\mathfrak{g})$ is a Poisson algebra. In particular: Your algebra $U$ has a canonical Poisson product $\{,\}$ on the symmetric algebra $Sym^*_K(U)$. Is this the product you define above?
Example: The tensor algebra $T(\mathfrak{g})$ of a $k$-Lie algebra $\mathfrak{g}$ is always a poisson algebra. There is a canonical quotient map $T(\mathfrak{g}) \rightarrow Sym^*_k(\mathfrak{g})$ and I believe this map is a map of Poisson algebras. Hence you get in your situation a canonical map
$$p: T(U) \rightarrow Sym^*_K(U) \cong K[x_1,x_2,x_3]$$
and you may compare the canonical Poisson product on $T(U)$ with the canonical Poisson product on $K[x_1,x_2,x_3]$. The universal enveloping algebra $\mathcal{U}(U)$ is a quotient of $T(U)$, and you may check if this gives a Poisson product on $\mathcal{U}(U)$.
Comment: "You are right that this rule give a structure of Poisson algebra on the symmetric algebra which is isomorphic to $K[x_1,x_2,x_3]$, but can we use this rule in $U$. Notice that $U$ is not commutative. I will add example to make my question more clear. – LearnMath928"
Answer: @LearnMath928 - If $\mathfrak{g}$ is a k-Lie algebra it follows the tensor algebra $T(\mathfrak{g})$ is canonically a Poisson algebra. The enveloping algebra $U:=\mathcal{U}(\mathfrak{g})$ is the quotient of the tensor algebra by the 2-sided ideal $I$ generated by elements on the form $x\otimes y -y\otimes x -[x,y]$. If $I$ is a 2-sided Poisson ideal it follows the quotient $U:=T(\mathfrak{g})/I$ is canonically a (non-commutative) Poisson algebra. You must check if $I$ is a 2-sided Poisson ideal:
$$\{T(\mathfrak{g}), I\} \subseteq I, \{I ,T(\mathfrak{g}) \} \subseteq I  $$
This is an "elementary" calculation.
Example: If $k$ is a commutative ring and $V$ is a left $k$-module it follows the associative ring $R:=End_k(V)$ has a bracket $[f,g]:=f\circ g-g\circ f$ and $R$ is a $k$-Lie algebra. Moreover
$$ [f,g\circ h]=g \circ [f,h]+[f,g] \circ h$$
for any $f,g,h\in R$. This holds for any associative $k$-algebra $R$: The bracket product $[x,y]:=xy-yx$ satisfies $[x,yx]=y[x,z]+[x,y]z$.
Comment: "Thank you, but this is just general information that will not help me to know the answer (this rule is correct or it is not correct for our algebra U). For example if you mean that this rule is correct because of the information you gave, then this rule would be correct in the enveloping algebra of any simple Lie algebra. But I think it is clear that this rule is not correct for example in the enveloping algebra $U(sl(3))$. – LearnMath928 11 hours ago"
Answer: By the above argument it follows the enveloping algebra $U:=\mathcal{U}(\mathfrak{g})$ of any $k$- Lie algebra $\mathfrak{g}$ has a bracket product $\{,\}$ where
$$\{u,v\}:=u \circ v - v \circ u$$
where $\circ $ is the product in $U$. This product gives $U$ the structure of a $k$-Lie algebra and the following holds:
$$\{u,v \circ w\}= u \circ v \circ w - v \circ w \circ v =$$
$$ v \circ (u \circ w- w \circ u) + (u \circ v - v \circ u) \circ w =$$ $$ v \circ \{u,w\}+\{u,v\} \circ w.$$
You must relate this to the Poisson product(s) on the tensor algebra. In your question you have chosen a basis/coordinates and it will simplify your study if you instead define your product intrinsically in terms of the Lie structure $[,]$ on $\mathfrak{g}$.
Note: You find the following statement on wikipedia: (Any) Poisson bracket $\{,\}$ (on a manifold $M$) is given locally as follows:
$$\{f,g\}:= \sum_{i,j} \{x^j, x^j\} \frac{\partial f}{\partial x^i} \frac{\partial g}{\partial x^j}$$
where $x^i$ are local coordinates. This is similar to your product defined above. This is a "commutative Poisson product". You want a "similar product" in the enveloping algebra. If $\mathfrak{g}$ has $k$-basis $x_1,..,x_n$ it follows the symmetric algebra $Sym^*_k(\mathfrak{g}) \cong k[x_1,..,x_n]$ is the polynomial ring and here you can take partial derivatives. The universal enveloping algebra $U:=\mathcal{U}(\mathfrak{g})$ has a PBW basis $x_1^{p_1}\cdots x_n^{p_n}$ with $p_i\geq 0$ but it is not clear what you mean by taking partial derivatives in $U$. The Poisson product
is a derivation  and you get formulas such as
$$\{x,y^2\}=2y[x,y]+[[x,y]y]=2y\{x,y\}+\{[x,y],y\}.$$
Question: "Does this hold in U? where ∂f∂xi is the standard derivetive of f with respect to xi."
Note: If $A:=Sym^*_k(\mathfrak{g})\cong k[x_1,..,x_n]$ is a polynomial ring over a field $k$ (of characteristic zero for simplicity), you may define the module of $k$-linear derivations $Der_k(A)$, and it follows the partial derivatives $\partial_{x_i} \in Der_k(A)$ are derivations.  In your formula you have written down a formula using partial derivatives $\frac{\partial f}{\partial x_i}$ where you view the variables $x_i$ and the element $f$ in the enveloping algebra $U$ and it is not clear that this can be done.
