Note that $$\lim_{x\to 0}\frac{\ln(1+x)}{x}=1\quad\text{and}\quad\lim_{x\to 0}\frac{e^x-1}{x}=1$$

We have $$\lim_{x\to\pi}\frac{\ln(2+\cos{x})}{\left(3^{\sin{x}}-1\right)^2}=\lim_{x\to\pi}\frac{\ln(1+(1+\cos{x}))}{\left(e^{\sin{x}\ln{3}}-1\right)^2}=\lim_{x\to\pi}\frac{1+\cos{x}}{\sin^2{x}\ln^2{3}}=\frac{1}{\ln^2{3}}\lim_{x\to\pi}\frac{-\sin{x}}{2\sin{x}\cos{x}}=\boxed{\frac{1}{2\ln^2{3}}}$$