Extrema of function - What is $\lambda$? We have the function $f(x_1, x_2, x_3)=9x_1\cdot x_2\cdot x_3$ and we want to find possible extremas under the constraint $2x_1+x_2+x_3=m, m>0$ and $x_1, x_2, x_3>0$. 
I have done the following: 
\begin{equation*}2x_1+x_2+x_3=m \Rightarrow x_3=m-2x_1-x_2\end{equation*} 
\begin{equation*}\tilde{f}(x_1, x_2)=9x_1\cdot x_2\cdot (m-2x_1-x_2)=9mx_1\cdot x_2-18x_1^2\cdot x_2-9x_1\cdot x_2^2\end{equation*} 
\begin{align*}\tilde{f}_{x_1}=9m x_2-36x_1\cdot x_2-9 x_2^2 \\ \tilde{f}_{x_1x_1}=-36x_1\cdot x_2 \\ \tilde{f}_{x_2}=9mx_1-18x_1^2-18x_1\cdot x_2 \\ \tilde{f}_{x_2x_2}=-18x_1 \\ \tilde{f}_{x_1x_2}=9m -36x_1-18 x_2\end{align*} 
\begin{equation*}\tilde{f}_{x_1}=0 \Rightarrow 9m x_2-36x_1\cdot x_2-9 x_2^2=0\Rightarrow x_2=0 \text{ or } x_2=m-4x_1\end{equation*} 
\begin{equation*}\tilde{f}_{x_1}=0 \Rightarrow 9m x_2-36x_1\cdot x_2-9 x_2^2=0\Rightarrow x_2=0 \text{ or } x_2=m-4x\end{equation*} 
So we get the following: 
\begin{equation*} \tilde{f}_{x_1x_1}(x_1, m-4x_1)=-36x_1\cdot (m-4x_1), \tilde{f}_{x_2x_2}(x_1,m-4x_1)=-18x_1, \tilde{f}_{x_1x_2}(x_1,m-4x_1)=9m -36x_1-18 (m-4x_1)\end{equation*} 
So, it holds that \begin{equation*}\tilde{f}_{x_1x_1}(x_1, m-4x_1)\tilde{f}_{x_2x_2}(x_1, m-4x_1) - \tilde{f}^2_{x_1x_2}(x_1, m-4x_1)=18\cdot 36x_1^2(m-4x_1)-(9m -36x_1-18 (m-4x_1))^2\end{equation*} 
For some specific  $m$ it is $>0$, then the function has extrema, right? 
$$$$ 
Then we have to calculate $x_1^{\star}(m), x_2^{\star}(m), \lambda^{\star}(m)$. 
What is $\lambda$ ? 
 A: Firstly, observe that the function has no minimum. It can go as close to $0$ as you want but never reaches $0$.
Secondly, see that $x_2$ and $x_3$ are symmetric. So the AMGM inequality says, fixing any $x_1$, the maximum is given by $x_2=x_3$. 
Therefore, you are left with the following problem: maximize $9x_1 \times y^2$ under the constraint $2x_1+2 y=m$ (for we have set $x_2=x_3=y$). At this stage, the problem can be solved with single-variable calculus by substituting the constraint to the objective function.
A: The fact that your problem mentions $\lambda$ says to me that Lagrange multipliers are the preferred solution method.  Let $g(x_1,x_2,x_3) = 2x_1 + x_2 + x_3$.  You are asked to find the maximum of $f$ on the set constrained by $g=m$, $x_1>0$, $x_2>0$, $x_3>0$.  The Lagrange multiplier equations are $\nabla f = \lambda \nabla g$, or
\begin{align}
    9x_2 x_3 &= 2\lambda \tag{1}\label{1} \\
    9x_1 x_3 &= \lambda \tag{2}\label{2} \\
    9x_1 x_2 &= \lambda \tag{3}\label{3} \\
\end{align}
along with the constraint equation
$$
    2x_1 + x_2 + x_3 = m \tag{4}\label{4}
$$
Comparing the left-hand sides of \eqref{2} and \eqref{3}, we get $9x_1x_3 = 9x_1 x_2$.  Since $x_1 > 0$, we can cancel it and get $x_2 = x_3$.  Similarly, comparing the left-hand sides of \eqref{1} and \eqref{2}, we get $9x_2 x_3 = 2(9x_1 x_3)$, so $x_2 = 2x_1$.  Hence $x_3 = 2x_1$ as well.  Substituting these into \eqref{4} gives 
$$
    2x_2 + 2x_1 + 2x_1 = m \implies x_1 = \frac{m}{6}
$$
and therefore $x_2 = x_3 = \frac{m}{3}$.  In the notation of your problem:
\begin{align*}
    x_1^\star(m) &= \frac{m}{6} \\
    x_2^\star(m) &= \frac{m}{3} \\
    x_3^\star(m) &= \frac{m}{3} \\
    \lambda^\star(m) &= 9 x_1^\star(m) x_2^\star(m) = 9\cdot \frac{m}{6} \cdot\frac{m}{3} = \frac{m^2}{2}
\end{align*}
You asked:

What is $\lambda$?

and I think you meant, what is $\lambda$ in terms of $m$ in this problem?  You have your answer to that.  But the deeper question is whether this seemingly arbitrary variable inserted into the solution method has any meaning.  And it does.
Let $f^\star(m)$ be the constrained maximum value of $f$, as a function of $m$.  We know
$$
    f^\star(m) = f(x_1^\star(m),x_2^\star(m),x_3^\star(m))
    = 9 \cdot \frac{m}{6}\cdot \frac{m}{3} \cdot \frac{m}{3} = \frac{m^3}{6}
$$
Notice that
$$
    \frac{df^\star}{dm} = \frac{m^2}{2} = \lambda
$$
This is true in general.  In words, the Lagrange multiplier is the change in the constrained maximum as the constraint level changes.  In economics, a lot of times $f$ is a utility function and $g$ is a cost function.  Then $\lambda$ is the marginal utility of money.
