The expression simplifies to $$\lim_{x\to 3^-}\dfrac{(27^{x+3})^{x/27} -9}{3^x-27}=\lim_{h\to 0} \frac{27^{(6-h)(3-h)/27} -9}{3^{3-h}-27}.$$ I simplified the denominator as follows: $\;27(3^{-h}-1)=-27h \ln 3$.
How should I simplify the numerator now?
On
The first thing I should do is to let $x=3-y$; simplifying, this gives $$\frac{3^y-3^{\frac{y^2}{9}}}{3 \left(3^y-1\right)}$$ Now, write $3^k=e^{k \log(3)}$ and use Taylor expansion around $k=0$ to get $$f(k)=e^{k \log(3)}=1+k \log (3)+\frac{1}{2} k^2 \log ^2(3)+O\left(k^3\right)$$ So, now compute $$\frac 13\frac {f(y)-f\left(\frac{y^2}{9}\right)}{f(y)-1}$$
On
You can also use the rule , $\lim_{h\to 0 } \frac{a^h - 1}{h} = \ln a $, simplifying we get $\lim_{h\to 0} \frac{9 (3^{\frac{-9h+ h^2}{9}} -1)}{-9h+h^2} \frac{-h}{27(3^{-h} - 1)} \frac{-9h+h^2}{-h} $ , now after applying the rule $\lim_{h\to 0 } \frac{a^h - 1}{h} = \ln a $, you can get your required limit.
On
$$L=\lim_{h\to 0} \frac{3^{(6-h)(3-h)/9}-9}{3^{3-h}-27} \to \frac{0}{0}$$ Apply L'Hospital's rule. $$L=\lim_{h \to 0} \frac{3^{(6-h)(3-h)/9}(-9+2h)/9}{3^{3-h}(-1)}=\frac{1}{3}.$$
On
Just to give a slightly different approach (albeit still along the lines of Harish Chandra Rajpoot's answer), let $x=3v+3$. Then
$${(27^{x+3})^{x/27}-9\over 3^x-27}={3^{v^2+3v+2}-9\over3^{3v+3}-27}={9(3^{v^2+3v}-1)\over27(3^{3v}-1)}={1\over3}\left(3^{v^2+3v}-1\over v^2+3v\right)\left(3v\over3^{3v}-1 \right)\left(v^2+3v\over3v \right)$$
Now
$$\lim_{v\to0}\left(3^{v^2+3v}-1\over v^2+3v\right)=\lim_{v\to0}\left(3^{3v}-1\over3v \right)=\lim_{u\to0}\left(3^u-1\over u \right)=\ln3$$
(although it really only matters that the two limits with $v\to0$ are equal and nonzero) and, obviously,
$$\lim_{v\to0}\left(v^2+3v\over3v \right)=1$$
Thus
$$\lim_{x\to3}{(27^{x+3})^{x/27}-9\over 3^x-27}=\lim_{v\to0}{9(3^{v^2+3v}-1)\over27(3^{3v}-1)}={1\over3}\cdot\ln3\cdot{1\over\ln3}\cdot1={1\over3}$$
(Note, there's no need to restrict to the one-sided limit.)
$$\lim_{h\to 0}\frac{27^{\frac{(6-h)(3-h)}{27}}-9}{3^{3-h}-27}=\lim_{h\to 0}\frac{27^{\frac{18-h^2-9h}{27}}-9}{27(3^{-h}-1)}$$ $$=\lim_{h\to 0}\frac{\left(9\cdot 27^{\frac{-h^2-9h}{27}}-9\right)}{27(3^{-h}-1)}$$ $$=\frac13\lim_{h\to 0}\frac{\left( 27^{\frac{-h^2-9h}{27}}-1\right)}{(3^{-h}-1)}$$ $$=\frac13\lim_{h\to 0}\frac{\left( 27^{\frac{-h^2-9h}{27}}-1\right)}{\frac{-h^2-9h}{27}}\cdot \frac{(-h)}{3^{-h}-1}\cdot \frac{\frac{h^2+9h}{27}}{h}$$ $$=\frac13\cdot \ln 27\cdot \frac{1}{\ln 3}\cdot \frac13$$ $$=\frac{3\ln3 }{9\ln 3}=\color{blue}{\frac13}$$