Let
$$ S_X^2 = \frac{1} {n} \sum_{i=1}^n X_i^2, S_Y^2 = \frac{1} {n} \sum_{i=1}^n Y_i^2$$
be the sample variance estimator of the variances $\sigma_X^2$ and $\sigma_Y^2$ respectively.
By the bivariate CLT,
$$ \sqrt{n} \left(\begin{bmatrix} S_X^2 \\ S_Y^2 \end{bmatrix} -
\begin{bmatrix} \sigma_X^2 \\ \sigma_Y^2 \end{bmatrix} \right)
\stackrel {d} {\to} \mathcal{N}
\begin{bmatrix} Var[X_1] & 0 \\ 0 & Var[Y_1] \end{bmatrix} $$
where
$$ Var[X_1] = E[X_1^4] - E[X_1^2]^2 = E[X_1^4] - \sigma_X^4$$
$$ Var[Y_1] = E[Y_1^4] - E[Y_1^2]^2 = E[Y_1^4] - \sigma_Y^4$$
Next we will need the bivariate Delta method to obtain the asymptotic distribution of $\displaystyle \frac {S_X^2} {S_Y^2}$
See, e.g. https://en.wikipedia.org/wiki/Taylor_expansions_for_the_moments_of_functions_of_random_variables
Consider $\displaystyle f(x, y) = \frac {x} {y}$. Then
$$ \frac {\partial f} {\partial x} = \frac {1} {y}, ~~
\frac {\partial f} {\partial y} = - \frac {x} {y^2}, ~~
\frac {\partial^2 f} {\partial x^2} = 0, ~~
\frac {\partial^2 f} {\partial y^2} = \frac {2x} {y^3}, ~~
\frac {\partial^2 f} {\partial x \partial y} = - \frac {1} {y^2}$$
We Taylor Expand $\displaystyle \frac {S_X^2} {S_Y^2} = f(S_X^2, S_Y^2)$ about the mean $(\sigma_X^2, \sigma_Y^2)$ up to the second order:
$$ \begin{align} f(S_X^2, S_Y^2) \approx &~ f(\sigma_X^2, \sigma_Y^2)
+ \left.\frac {\partial f} {\partial x}\right|_{(\sigma_X^2, \sigma_Y^2)}
(S_X^2 - \sigma_X^2)
+ \left.\frac {\partial f} {\partial y}\right|_{(\sigma_X^2, \sigma_Y^2)}
(S_Y^2 - \sigma_Y^2) \\
&~ + \frac {1} {2} \left.\frac {\partial^2 f} {\partial x^2}\right|_{(\sigma_X^2, \sigma_Y^2)}(S_X^2 - \sigma_X^2)^2
+ \frac {1} {2} \left.\frac {\partial^2 f} {\partial y^2}\right|_{(\sigma_X^2, \sigma_Y^2)}(S_Y^2 - \sigma_Y^2)^2 \\
&~ + \left.\frac {\partial^2 f} {\partial x \partial y}\right|_{(\sigma_X^2, \sigma_Y^2)}(S_X^2 - \sigma_X^2)(S_Y^2 - \sigma_Y^2) \\
\Rightarrow \frac {S_X^2} {S_Y^2} \approx &~ \frac {\sigma_X^2} {\sigma_Y^2}
+ \frac {1} {\sigma_Y^2} (S_X^2 - \sigma_X^2)
- \frac {\sigma_X^2} {\sigma_Y^4} (S_Y^2 - \sigma_Y^2) \\
&~ + 0 + \frac {\sigma_X^2} {\sigma_Y^6} (S_Y^2 - \sigma_Y^2)^2
- \frac {1} {\sigma_Y^4} (S_X^2 - \sigma_X^2)(S_Y^2 - \sigma_Y^2)
\end{align} $$
Taking expectation,
$$ E\left[\frac {S_X^2} {S_Y^2}\right]
\approx \frac {\sigma_X^2} {\sigma_Y^2} + 0 + 0
+ \frac {\sigma_X^2} {\sigma_Y^6} Var[S_Y^2] - 0
= \frac {\sigma_X^2} {\sigma_Y^2}
+ \frac {\sigma_X^2(E[Y_1^4] - \sigma_Y^4)} {n\sigma_Y^6} $$
and variance (with first order only)
$$ Var\left[\frac {S_X^2} {S_Y^2}\right]
\approx 0 + \frac {1} {\sigma_Y^4}Var[S_X^2] + \frac {\sigma_X^4} {\sigma_Y^8}Var[S_Y^2] = \frac {E[X_1^4] - \sigma_X^4} {n\sigma_Y^4}
+ \frac {\sigma_X^4(E[Y_1^4] - \sigma_Y^4)} {n\sigma_Y^8}$$
So by Delta method, collecting the terms,
$$ \sqrt{n} \left(\frac {S_X^2} {S_Y^2} - \frac {\sigma_X^2} {\sigma_Y^2}
\right) \stackrel {d} {\to} \mathcal{N} \left(0,
\frac {E[X_1^4] - \sigma_X^4} {\sigma_Y^4}
+ \frac {\sigma_X^4(E[Y_1^4] - \sigma_Y^4)} {\sigma_Y^8} \right)$$
The second order term here just serve as a correction term providing higher accuracy if needed.