Step 1: Spot the product.
We differentiate $y = x\,\log_e x$, a product of $x$ and $\log_e x$, so the product rule fits.
Step 2: Write the rule.
For $y = uv$, \[ \frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} \] with $u=x$ and $v=\log_e x$.
Step 3: Differentiate each piece.
Here $\dfrac{du}{dx} = 1$ and $\dfrac{dv}{dx} = \dfrac{1}{x}$.
Step 4: Plug in and tidy.
\[ \frac{dy}{dx} = x\cdot\frac{1}{x} + \log_e x \cdot 1 = 1 + \log_e x \]
Step 5: Answer.
\[ \boxed{1 + \log_e x} \]