Define $$Y_n = \frac{X_1 +X_2 + \cdots X_n}{X_1^2 +X^2_2 + \cdots X^2_n}\,,$$ where $X_1 ,X_2 , \dots$ are iid $U(-1,1)$.
Show that $Y_n$ when suitably normalized, converges in distribution to some distribution.
$\textbf{My Take:}$ Let $Z_n=X_1 +X_2 + \dots X_n$ and $W_n=X_1^2 +X_2^2 + \dots X_n^2$
By CLT, $\sqrt{n}(\frac {Z_n}{n} - E(X_1)) = \sqrt{n}(\frac{Z_n}{n})\stackrel{d}{\rightarrow}N(0, 1/3)$ since $E(X_1) = 0$ and $\sigma^2_X = 1/3$.
Similarly, $\sqrt{n}(\frac {W_n}{n} - E(X^2_1)) = \sqrt{n}(\frac{W_n}{n} - 1/3)\stackrel{d}{\rightarrow}N(1/3, 4/45)$, since $E(X^2_1) = 1/3$ and $\sigma^2_X = 4/45$.
Thus, $Y_n = \frac{N(0,1/3)}{N(1/3,4/45)}$ I am stuck-in here. I don't know how to simplify this.
$\textbf{Textbook solution:}$ The numerator is $N(0,1/3)$ and the denominator is $1/3$ and so $\sqrt{n}Y_n = \frac{\sqrt{n} Z_n/n}{W_n/n}\stackrel{d}{\rightarrow} \frac{N(0,1/3)}{1/3} = 3N(0,1/3) = N(0,3)$
$\textbf{Questions}$
Why is my approach wrong?
If my approach is not wrong, how do I simplify the ratio of two distributions?
From the textbook solution, what is the algebra from $3N(0,1/3)$ to $N(0,3)$ i.e. the trick.
Your approach doesn't quite work because $Z_n$ and $W_n$ are not independent sequences of random variables. You can instead use the Law of Large Numbers to show that $W_n/n \rightarrow 1/3$ in probability. Finally invoke Slutsky's Theorem as stated in the comments.