This doesn't really differ from the logic of the answers posted by Alex Zorn and mixedmath, but the presentation might carry some "weight."
Let's label the balls $N$ and $n$ for New York, $C$ and $c$ for California, and $T$ and $t$ for Texas. Our job is to partition these into a "heavy" set $H$ and a "light" set $L$ with the uppercase and lowercase going into different sets for each letter.
For the first weighing, compare $N$ and $C$ to $n$ and $T$. It's convenient to think of the labels as weights and express the outcome as $N+C\approx n+T$, where "$\approx$" is either $>$, $=$, or $<$. By an elementary relabeling symmetry (swapping the labels $N$ and $n$), it suffices to consider only the first two cases.
If $N+C\gt n+T$, we can conclude that $N\in H$ and $C\ge T$. This implies $H$ is one of three possibilities: $\{N,C,T\}$, $\{N,C,t\}$, or $\{N,c,t\}$. We can determine which by comparing $c$ and $T$ in the second weighing.
If $N+C=n+T$, then $N$ and $C$, and likewise $n$ and $T$, must be in different sets, which we can express as
$$\{H,L\}=\{\{N,T,c\},\{n,t,C\}\}$$
It's clear from this that we can determine which set on the right hand side is $H$ and which is $L$ by simply comparing, say $N$ and $n$.