How to evaluate $\lim _{x\to \infty }\left(\frac{x+3}{\sqrt{x^2-5x}}\right)^{x^2\sin\left(1/x\right)}$?

You're making your own life more difficult. ;-) But your idea is good.

After taking the logarithm, apply the substitution $x=1/t$ where it's not restrictive to assume $x>0$ (actually, $x>5$); note that $$ \frac{x+3}{\sqrt{x^2-5x}}=\frac{1+3t}{\sqrt{1-5t}}, $$ so you have $$ \lim_{t\to0^+}\frac{\sin t}{t^2}\ln\frac{1+3t}{\sqrt{1-5t}}= \lim_{t\to0^+}\frac{\ln(1+3t)-\frac{1}{2}\ln(1-5t)}{t} $$ owing to $\lim_{t\to0}\frac{\sin t}{t}=1$ (of course, conditionally to the existence of the last limit).

This can be rewritten $$ 3\lim_{t\to0^+}\frac{\ln(1+3t)}{3t}+ \frac{5}{2}\lim_{t\to0^+}\frac{\ln(1-5t)}{-5t}=3+\frac{5}{2} $$ or with Taylor up to degree $1$, $$ \lim_{t\to0^+}\frac{3t+\frac{5}{2}t+o(t)}{t}=3+\frac{5}{2} $$