Read-off particle from (projected) Dynkin labels In the review of Slanksy "Group theory for unified model building" in chapter 6:
How do one relate the projected Dynkin diagrams from for example $\overline{5}+10$ of $su(5)$ to the corresponding representation in of the subgroup $su(2)\times su(3)$?
An example is 
$P( 0 0 1)=(0)(01)$, how do one see that this must be a charge $1/3$ antiquark singlet?
Reference: http://cds.cern.ch/record/134739/files/198109187.pdf?origin=publication_detail
 A: The first set of parentheses contains the $SU(2)$ weight. It being $(0)$ tells you that you have an $SU(2)$ singlet. The second set of parentheses is the rep under the $SU(3)$ part. 
Looking at dynkin labels of different reps you find that $(01)$ belongs to the antifundamental representation of $SU(3)$.
Finally there is the question of the $U(1)$ charge. This is not uniquely defined in the notation you gave. So $(0)(01)$ can either be a charge $-1/3$ up antiquark or charge $2/3$ down antiquark.
The $U(1)$ charges are uniquely fixed once you identify one single state and define its $U(1)$ charge. Here, one could for example take the positron. It has $(0)(00)$ and therefore is unique in $\overline{ \mathbf 5} + \mathbf{10}$. Knowing it has Hypercharge $Y = 1$ allows you to define a linear combination of operators (corresponding to a certain sum of dynkin indices) as the Hypercharge and use this to determine all other $U(1)$ charges.
EDIT to adress comment:
Let's look at the representation with $(01)$ as a highest root (which is the antifundamental). From that we can subtract a simple root $(-1 2)$ to get
$$ (01) \to (1 -1)$$
from which we can subract the other simple root $(2 -1)$ to get
$$ (1 -1) \to (-1 0)$$
Having only negative or 0 entries this is the bottom of the rep. Since there was no branching we can summarize
$$ (0 1) \overset{\alpha_2}{\to} (1 -1) \overset{\alpha_1}{\to} (-1 0 )$$
We see we have found a three dimensional rep. Now there is another three dimensional rep, the fundamental $(1 0)$. Which of these we call fundamental and which antifundamental is arbitrary, since they swap roles in the group homomorphism $T \to T^*$ (which, iirc corresponds to the non-trivial symmetry of the Dynkin diagram).
So once we set the rep starting form $(1 0)$ as the fundamental, the rep starting from $(0 1)$ is the antifundamental, containing the states 
$$\overline{\mathbf 3} = {(0 1), (1 -1), (-1 0)}$$
and finding $(0 1)$ in that list makes us identify it as a member of $\overline{\mathbf 3}$.
P.S.: We could also have seen that $\overline{\mathbf 3}$ and $\mathbf 3$ are conjugate from seeing that if $(a b)$ is a member of $\overline{\mathbf 3}$, $(-a -b)$ is a member of $\mathbf 3$!
