In the last line of this Wikipedia article on representations of sl_3 there is the line: "There is also a simple pattern to the multiplicities of the various weight spaces. Finally, the irreducible representations with highest weight (0,m) can be realized concretely on the space of homogeneous polynomials of degree m in three complex variables." With a reference to Brian Hall's book Exercise 6.8. I checked this exercise and honestly it's not even about $\mathfrak{sl}_3$ but $\mathfrak{sl}_2$, so I think I am not looking at the correct exercise.
To start my work on this, I know that the Cartan subalgebra $\mathfrak{h}$ is spanned by $H_1 = \begin{pmatrix}1 & 0 & 0\\0&-1&0\\0&0&0\end{pmatrix}$ and $H_2 = \begin{pmatrix}0 & 0& 0\\0&1&0\\0&0&-1\end{pmatrix}$. In the case of degree 2 (let's called the polynomial space $Sym^n(\mathbb{C}^3)$ as $S_n$) I know that I can take the action of $H_1$ and $H_2$ on the basis and recover the "weights" or eigenvalues.
For example: $H_1(x^2) = H_1(x)x + x H_1(x) = 2x^2$ so the eigenvalue is 2. I can probably extend this process to see the action of $H_i$ on higher powers to get a complete description of what the eigenvalues are (although I'm currently not sure what that is).
So some of my questions are:
How can I even be sure this $S_n$ is a representation of $\mathfrak{sl}_3$? Does something about the behavior of the Cartan subalgebra on $S_n$ imply this? Does it also imply that it is an irreducible representation?
How does the claim from Wikipedia actually work? As in, how can you realize irreducible reps with highest weight $(0,m)$ in $S_n$?
Wikipedia is just jumping straight to the result of course but there are a number of ways you could show this. I'm going to take a fairly general approach so you can use this to see the big picture. To be clear, the space of homogenous polynomials is exactly the irreducible representation of highest weight $(0,m)$.
Firstly, to answer your first question you can just construct a valid representation on $S_n$. Perhaps the best way to do this is show how to build representations using tensor products. Let $\rho_V$ and $\rho_W$ be two representations of a Lie algebra on $V$ and $W$. Then $V\otimes W$ is a representation where the action is $\rho_{V\otimes W}(X) = \rho_{V}(X)\otimes I + I\otimes \rho_{W}(X)$. You can check this defines a representation but this comes from differentiating the natural "diagonal" Lie group representation using product rule.
This works equally well on symmetric and alternating tensor powers. For example $X(v\wedge w) := X(v) \wedge w + v \wedge X(w)$ and $ X(vw) = X(v)w + vX(w) $
So $S^n(V)$ and $\bigwedge^n(V)$ are also representations if $V$ is. Indeed you can think of these as subspaces of the tensor power $\bigotimes^n V = V\otimes \dotsm \otimes V$ and the fact that they are subrepresentations follows from the action commuting with permuting the order of the tensors (these ideas in full lead to Schur-Weyl duality).
Now $S_n$ the space of homogeneous polynomials of degree $n$ on $V$ is precisely $S^n(V^*)$ and we have a representation on that space by combining the above with the dual (aka contragredient) representation on $V^*$: $X(f) := - f \circ X$.
To see that this representation is irreducible you can simply show that you can get from the span of any polynomial to the span of any other by repeated application of elements of $\mathfrak{sl}_3$. I advise considering how the root vectors act.
To see that this is exactly the representation given by $(0,m)$ we first note that the tensor product of two representations has highest weight the sum of the two highest weights (indeed all of its weights are sums of the original weights). The dual representation has highest weight $(0,1)$ so $\bigotimes^m V^*$ has highest weight $(0,m)$ and indeed that highest weight vector is given by $v\otimes \dotsm \otimes v$ where $v$ is a highest weight vector in $V^*$ but this is in $S^mV^*$ so that is the irreducible representation with that highest weight. Alternatively you could just directly show that this is the highest weight in $S^mV^*$ and by the Theorem of the Highest Weight it is the unique irreducible representation with that highest weight.
The reason for venturing into tensor products is that all irreducible representations of $\mathfrak{sl}_3$ can be viewed as subrepresentations of some tensor product of $\mathbb{C}^3$ (even the dual is simply $(\mathbb{C}^3)^* \cong \bigwedge^2 \mathbb{C}^3$)