= Solution
Convergence in distribution makes $(X_n)$ tight. Hence for every $\eta>0$ there is $M$ such that $\mathbb P(|X_n|>M)<\eta$ for all sufficiently large $n$. Since $Y_n-c\xrightarrow{P}0$,
$$
\mathbb P(|X_n(Y_n-c)|>\varepsilon)
\leq
\mathbb P(|X_n|>M)
+\mathbb P(|Y_n-c|>\varepsilon/M)
\longrightarrow\leq\eta.
$$
Letting $\eta\downarrow0$ gives $X_n(Y_n-c)\xrightarrow{P}0$. Also, the continuous-mapping theorem gives $cX_n\xrightarrow{d}cX$. Since
$$
X_nY_n=cX_n+X_n(Y_n-c),
$$
the <Slutsky theorem> yields
$$
\boxed{X_nY_n\xrightarrow{d}cX}.
$$
Now write the requested statistic as
$$
\frac{n^{-1/2}\sum_{i=1}^nZ_i}
{n^{-1}\sum_{i=1}^nZ_i^2}.
$$
The <central limit theorem> gives
$$
n^{-1/2}\sum_{i=1}^nZ_i\xrightarrow{d}N(0,1).
$$
Since $\mathbb E[Z_i^2]=1<\infty$, the <strong law of large numbers> gives
$$
\frac1n\sum_{i=1}^nZ_i^2\longrightarrow1
$$
almost surely, and therefore its reciprocal converges in probability to one. Applying the product result proves
$$
\boxed{
\frac{\sqrt n\sum_{i=1}^nZ_i}{\sum_{i=1}^nZ_i^2}
\xrightarrow{d}N(0,1)
}.
$$
This is a <self-normalized central limit theorem with a second-moment denominator>.
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