Prove the Droid Trader Problem is NP-complete
You can find a simple reduction from Knapsack or from the Partition Problem (which itself reduces to Knapsack).

Let's do it from the Partition Problem:


  Given $n$ integers $S = \{a_1, a_2\ldots, a_n\}$ with even total sum, does there exist a subset whose sum is exactly half the sum of all elements ?


So suppose you are given an instance of the Partition Problem. We construct an instance of the Droid Trader problem as follows:


There will be $n$ types of droids, one for each integer in $S$. There will be two planets $s$ and $t$, with a directed edge from $s$ to $t$.
On planet $s$, the player can buy a single droid of each type $i$ for a price of $a_i$, and can sell nothing.
On planet $t$, the player can sell droids of every type $i$ for a price of $2\cdot a_i$, and can buy nothing.
The player starts with $z$ equal to half the sum of all elements in $S$.


Then, it is easy to see that, the player can never end up with more money than $\sum S$. Moreover, he can achieve $\sum S$ if and only if he can buy $\frac{1}{2}\sum S$ worth of droids on the first planet, i.e. there exists a subset of $S$ with sum $\frac{1}{2}\sum S$.

So the original question on the Partition Problem instance is equivalent to asking if the player can end up with $\sum S$ money in the corresponding Droid Trader problem.

Because this is a polynomial time reduction, this proves the claim.