Given that, $ 2^{y^{2} \log_{3}5} = 5^{\log_{2}3}; y < 0 $
Taking $\log_{3}$ on both sides.
$ \log_{3} 2^{y^{2} \log_{3}5} = \log_{3} 5^{\log_{2}3} $
$ \Rightarrow y^{2} \log_{3}5 \; \log_{3}2 = \log_{2}3 \; \log_{3}5 $
$ \Rightarrow y^{2} \log_{3}2 = \log_{2}3 $
$ \Rightarrow y^{2} = \dfrac{\log_{2}3}{\log_{3}2} $
$ \Rightarrow y^{2} = \dfrac{\log_{2}3}{\frac{1}{\log_{2}3}} \quad \left[ \because \log_{a}b = \dfrac{1}{\log_{b}a}\right] $
$ \Rightarrow y^{2} = \log_{2}3 \cdot \log_{2}3 $
$ \Rightarrow y^{2} = \left( \log_{2}3 \right)^{2} $
$ \Rightarrow y = \; – \log_{2}3 \quad [\because y \;\text{is negative}]$
$ \Rightarrow y = \log_{2}3^{-1} $
$ \Rightarrow \boxed{ y = \log_{2} \frac{1}{3}} $
Correct Answer$: \text{A}$