1. Interpret the equation

Given $x\log_e(\log_e x)-x^2+y^2=4, \qquad y>0$

We need to find $\dfrac{dy}{dx}$ at $x=e$.

2. Differentiate implicitly

Differentiate both sides with respect to $x$: $\frac{d}{dx}\left[x\ln(\ln x)\right]-\frac{d}{dx}(x^2)+\frac{d}{dx}(y^2)=0$

Now, $\frac{d}{dx}\left[x\ln(\ln x)\right]=\ln(\ln x)+x\cdot \frac{1}{\ln x}\cdot \frac{1}{x}=\ln(\ln x)+\frac{1}{\ln x}$

Also, $\frac{d}{dx}(x^2)=2x, \qquad \frac{d}{dx}(y^2)=2y\frac{dy}{dx}$

So, $\ln(\ln x)+\frac{1}{\ln x}-2x+2y\frac{dy}{dx}=0$

Hence, $2y\frac{dy}{dx}=2x-\ln(\ln x)-\frac{1}{\ln x}$

Therefore, $\frac{dy}{dx}=\frac{2x-\ln(\ln x)-\frac{1}{\ln x}}{2y}$

3. Find $y$ at $x=e$

Substitute $x=e$ into the original equation: $e\ln(\ln e)-e^2+y^2=4$

Since $\ln e=1$, we get $\ln(\ln e)=\ln 1=0$

Thus, $0-e^2+y^2=4$ Rightarrow y^2=4+e^2$$

Given $y>0$, $y=\sqrt{4+e^2}$

4. Evaluate $\dfrac{dy}{dx}$ at $x=e$

At $x=e$, $\ln(\ln e)=0, \qquad \frac{1}{\ln e}=1$

So, $\left.\frac{dy}{dx}\right|_{x=e}=\frac{2e-0-1}{2\sqrt{4+e^2}}=\frac{2e-1}{2\sqrt{4+e^2}}$

5. Match with options

This matches: $\boxed{\text{C: } \frac{2e-1}{2\sqrt{4+e^2}}}$