How do I prove if this logical argument is valid using a truth table? Is the following logical argument valid?
$P$: You can not teach an old dog to sit
$Q$: Bill is a dog
$R$: You can not teach Bill to sit
$\longrightarrow$ C: Bill is old

How do I prove if the above logical argument is valid or not using a truth table?
 A: Your argument is invalid : you could have young dogs who can't be taught to sit. Indeed, your assumption P says nothing about young dogs.
A: If we interpret these statements using standard logical expressions, like conjunctions,  and implication, we can say (as already stated by Mauro in his comment):
$P:$ $Old(x) \land Dog(x) \implies \lnot TeachToSit(x)$
$Q:$ $Dog(Bill)$
$R:$ $\lnot TeachToSit(Bill)$
The truth table for implication can be a little tricky to understand, it is:
$$\begin{array}{|l |l| c|}
\hline
X&Y&X \implies Y\\
\hline
F&F&T\\
F&T&T\\
T&F&F\\
T&T&T\\
\hline
\end{array}$$
One way to think of it is that the only way to falsify an implication is to show that the condition (left-hand side) does not generate the outcome (right-hand side).

To explore the truth-value of the statement "Bill is old" ($Old(Bill)$) using truth tables, we simply write down all the possible combinations, given the facts that "Bill is a dog", and "You can not teach Bill to sit" and see which combinations satisfy $P$ .
So we only consider combinations where $Dog(Bill) = T$ and $TeachToSit(Bill) = F$:
$$\begin{array}{|c |c| c|c|}
\hline
Old(Bill)&Dog(Bill)&TeachToSit(Bill)&Old(Bill)\land Dog(Bill) \implies \lnot TeachToSit(Bill)\\
\hline
F&T&F&T\\
T&T&F&T\\
\hline
\end{array}$$
From this we see that the implication is satisfied either way, therefore we cannot tell if Bill is old or not, the truth-value of $Old(Bill)$ is unknown with only the information given. So the argument is not valid.

The constructive reason you cannot tell whether Bill is old or not is that implications are one-way, that's why the arrow points to the right. If $X \implies Y$, we say that $X$ is sufficient but not necessary for $Y$, and v.v. $Y$ is necessary but not sufficient for $X$, this wikipedia article goes into detail.
The layman's reason why you cannot argue as in your question is that (given the information you were given) old dogs aren't necessarily the only unteachable dogs, there can very well be unteachable dogs that aren't old.
