Find $\lim\limits_{x\to 0}\left (\frac{1^x+2^x+3^x+\dots+n^x}{n} \right)^{\frac1x}$

Actually, $\ln(y)=\frac{1}{x}\ln{\frac{(1^x+2^x+...+n^x)}{n}}$. Whenever after putting $x=0$ you get the form $1^{\infty}$, you need to take $\ln$ and find the limit.

Here, after putting $x=0$, we have $\big(\frac{1^0+2^0+\cdots +n^0}{n}\big)^{1/0}=1^{\infty}$. So taking $\ln$ in both side and finding limit is correct approach.

Define $S=\{k \mid k\in \mathbb{Z}^+ \land k\le n\}$. Define the limit as $L$. Rewrite it as $\exp \ln L$ and use L'Hopital's rule and you'll be done in no time.

$$\begin{aligned}\lim_{x\to 0}\left(\dfrac{\displaystyle\sum_{k\in S}k^x}{n}\right)^{1/x}&=\lim_{x\to 0} \exp \dfrac{1}{x}\ln\left(\dfrac{\displaystyle\sum_{k\in S}k^x}{n}\right)\\&=\lim_{x\to 0}\exp \dfrac{n}{\displaystyle\sum_{k\in S}k^x}\cdot\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{\displaystyle \sum_{k\in S} k^x}{n}\right]\\&=\lim_{x\to 0}\exp \dfrac{n}{\displaystyle \sum_{k\in S}k^x}\cdot\dfrac{\displaystyle\sum_{k\in S}k^x\ln k}{n}\\ &=\lim_{x\to 0}\exp \dfrac{\displaystyle\sum_{k\in S}k^x\ln k}{\displaystyle \sum_{k\in S}k^x}\to \exp \dfrac{\ln n!}{n}=\sqrt[n]{n!}\end{aligned}$$

Your basic approach is fine but there are some problems. The first is where you say "taking $\lim$ on both sides". You can't take the limit unless you know the limit exists. But that is part of the exercise, right? I would omit the $\lim$ business until the end. Second, why do you still have $\ln y$ on the left after "taking the limit"? Finally, you didn't get the correct expression in using LHR. It should be

$$\frac{n}{1^x+2^x+\cdots +n^x}\cdot\frac{1^x\ln(1)+\cdots +n^x\ln(n)}{n}.$$