Let's first express $x$ in terms of $y$ and then take derivative on both sides:

$$y=\ln{x}\implies x=e^y\implies d x=d \left(e^y\right)$$

Meanwhile, we have

$$\frac{d \left(e^y\right)}{dy}=e^y \implies d {\left(e^y\right)}=e^y d y$$

Setting this to the previous relation yields

$$d x=d \left(e^y\right) \implies d x=e^y d y\implies \frac{dy}{dx}=\frac{1}{e^y}$$

Noting $e^y=x$. Therefore, the desired result is

$$\frac{dy}{dx}=\boxed{\frac{d}{dx} \ln{x}=\frac{1}{x}}$$

Alternatively, this problem can also be solved using the inverse function rule.

In this case, $g(x)=\ln{x}$, $f(x)=e^x$, and $f'(x)=e^x$. Therefore

$$g'(x)=\frac{1}{f'(g(x))}=\frac{1}{e^{g(x)}}=\frac{1}{e^{\ln{x}}}=\frac{1}{x}$$