It is a remarkable and beautiful fact that the irreducible representations of the symmetric group $S_n$ are in correspondence with the partitions of $\lambda \vdash n$. For example, in the case of $S_3$, the irreducible partitions correspond to all the partitions of 3, namely
$$(3) \quad (1,1,1) \quad (2,1)$$
The full story is too long for this post, but details can be found in Chapter 4 of 'Representation Theory: A first course' by Fulton and Harris.
To see the decomposition of the regular representation into its irreducible components is most easily done via character theory. Let $\chi$ be the character of the left regular representation (that is, let $S_3$ act on $K[S_3]$ on the left); $\chi(\sigma)$ is the number of fixed points of the action of $\sigma$ on $K[S_3]$ (as these contribute to the trace of this action). It is plain to see that the only element in $S_3$ which fixes anything in $K[S_3]$ is the identity element $e$, and moreover, this fixes every element in $K[S_3]$. Thus we have
$$\chi(e) = |S_3|=6 \qquad \chi(\sigma)=0 \quad \forall \ \sigma \in S_3 \backslash \{e\}$$
Let $\chi_\lambda$ be the character of the irreducible representation of $S_3$ corresponding to the partition $\lambda \vdash 3$. It is a fact from character theory that the inner product of characters of a representation $A$ and an irreducible representation $B$, defined by,
$$\langle \chi_A, \chi_B \rangle = \frac{1}{|G|}\left( \sum_{g \in G} \chi_A(g)\chi_B(g) \right)$$
gives the multiplicity of $B$ in $A$. For your question then, we need to compute this inner product with the character $\chi_\lambda$. For this we only need to know one more fact: that $\chi_\lambda(e)$ is the dimension of the corresponding irreducible representation of $S_3$ corresponding to $\lambda$. We can now compute
$$\langle \chi, \chi_\lambda \rangle = \frac{1}{|S_3|}\left(\chi(e)\chi_\lambda(e) \right) = \frac{1}{6}(6\cdot \chi_\lambda(e)) = \chi_\lambda(e)$$
We see that the irreducible representation corresponding to $\lambda$ appears in the decomposition of the regular representation exactly the 'dimension of representation' number of times.
Here are the correspondences in your case:
$$\lambda = (3) \rightarrow V_0, \dim = 1$$
$$\lambda = (1,1,1) \rightarrow V_1, \dim = 1$$
$$\lambda = (2,1) \rightarrow V, \dim = 2$$
Therefore
$$k[S_3] = V_0 \oplus V_1 \oplus V^{\oplus 2}$$
as desired.
If this is new to you, then there are a lot of details to check here, all of which can be found in Fulton Harris. You should know that this story works for any $n$ and we have in general that
$$K[S_n] = \bigoplus_{\lambda \vdash n} V_\lambda^{\oplus \dim V_\lambda}$$
where $V_\lambda$ is the irreducible representation of $S_n$ corresponding to the partition $\lambda$ of $n$.
This is essentially @QiaochuYuan's point. If $V\neq 0$ then pick any non-zero vector $v\in V$. The $\mathbb{C}$-linear span of the vectors $\{gv|g\in G\}$ is a non-zero $G$-invariant subspace of $V$. As the set of vectors $\{gv|g\in G\}$ has cardinality $|G|$, their span will have dimension at most $|G|$.
Note this is only true under the assumption $V\neq 0$. That should be in the question somewhere - if not you should mention it to your instructor.
Best Answer
I will do the example for $d=1$, because you seem to be confused about what kind of work you have to do. When you know what you have to do perhaps you will be able to do the rest on your own, so I'll stick with the $d=1$ case which is the most easy one to give you an idea.
You have $6$ elements in $S_3$, namely $()$, $(12)$, $(13)$, $(23)$, $(123)$ and $(132)$. You are given a representation of $S_3$ over the vector space $V_1$ of homogeneous polynomials of degree $1$, which is a vector space (over $\mathbb C$? You didn't specify the ground field of the vector space ; this is a very important fact when treating representation theory!) A basis of $V_d$ is given by $\{x_1, x_2, x_3\}$ (I prefer this notation, you will see why in a second) because given this, we can say that if $\pi$ is a permutation of $\{1,2,3\}$, then $$ \pi \cdot p(x_1, x_2, x_3) = p(x_{\pi^{-1}(1)},x_{\pi^{-1}(2)}, x_{\pi(3)}) $$ or in other words, you "apply the permutation to the indices". Note : it might seem counter-intuitive at first, but if $p(x_1,x_2,x_3) = a_1 x_1 + a_2 x_2 + a_3 x_3$, then $$ \pi \cdot p(x_1,x_2,x_3) = a_{\pi^{-1}(1)} x_1 + a_{\pi^{-1}(2)}x_2 + a_{\pi^{-3}}x_3. $$ This is dangerous to forget if you write vectors of $V_d$ in terms of their coefficients, since it might lead you to think that $\pi \cdot (a_1,a_2,a_3) = (a_{\pi(1)},a_{\pi(2)},a_{\pi(3)})$, which is not the case. (Try applying $(123)$ to $x_1 = 1 x_1 + 0 x_2 + 0 x_3$ if you are not convinced.)
Now we know a trivial submodule of $V_1$ : for instance, the submodule generated by $x_1 + x_2 + x_3$. By Maschke's theorem, there exists a complement to this submodule which is also a submodule, so that they are in a direct sum ; in other words, you can find a basis of $V_d$ of the form $\{x_1 + x_2 + x_3, y_2, y_3\}$ such that $\langle y_2, y_3 \rangle$ is a submodule of $V_1$. In matrix terms, this means that over this new basis, your representation has this form. $$ \rho(\pi) = \begin{bmatrix} * & 0 & 0 \\ 0 & * & * \\ 0 & * & * \end{bmatrix} $$ Now it remains to show if the last block is irreducible (representations of degree $1$ cannot be reduced).
It is very hard to precisely give you what your teacher wants you to answer, because there are many tools to do this ; character theory, module theory, "doing things by hand" (which is computationally speaking very hard, mostly for $d=3$) but if you tell me more I guess I could try to help you more.
Hope that helps,