Let $S_n = \frac{1}{n}\sum_{i=1}^n X_i$, and $T_n = \frac{1}{n}\sum_{j=1}^nY_i$, where
The $X_i$ are iid, the $Y_i$ are iid (with a different law)
$X_i$, and $Y_i$ are dependent
For $i\neq j$, $X_i$ and $Y_j$ are independent.
Is there a central limit type result for $S_n^2 - T_n^2$?
If $X_i$ and $Y_j$ are dependent for $i=j$, but independent for $i\neq j$ we have an iid sample from bivariate distribution $Z_i=(X_i,Y_i)$. Then central limit theorem gives us
\begin{align} \sqrt{n}\left(\frac{1}{n}\sum_{i=1}^nZ_i -EZ_1\right)\xrightarrow{D}N(0,\Sigma) \end{align} with $\Sigma=cov(Z_1)$ and $\xrightarrow{D}$ indicating convergence in distribution. Note that
$$(S_n,T_n)=\frac{1}{n}\sum_{i=1}^nZ_i.$$
Now we can use delta method, which states that if $r_n(U_n-\theta)\xrightarrow{D}U$ for numbers $r_n\to\infty$, then $r_n(\phi(U_n)-\phi(\theta))\xrightarrow{D}\phi'_\theta(U)$. This statement can be found here. Delta method is also described here.
In our case now we have $r_n=\sqrt{n}$, $\phi(x,y)=x^2-y^2$, $U_n=(S_n,T_n)$ and $\theta=(EX_1,EY_1)$. We have
$$\phi_{\theta}'=(2EX_1,-2EY_1)$$
and
$$U=(U_1,U_2)\sim N(0,\Sigma)$$
Finally we get
\begin{align} \sqrt{n}\left(S_n^2-T_n^2-(EX_1)^2+(EY_1)^2\right)\xrightarrow{D} 2U_1EX_1-2U_2EY_1:=V \end{align}
Since we know that $(U_1,U_2)\sim N(0,\Sigma)$ we get that
$$V\sim N(0,4(EX_1,EY_1)\Sigma(EX_1,EY_1)').$$
Note that I basically redid the example after the theorem in the link.
The final answer then depends on $\Sigma=cov((X_1,Y_1))$, but this should be known to original poster.