Note that $$ \frac{1}{e} + x \log x \geq 0 $$ for all positive $x,$ and gives $0$ only at $x = \frac{1}{e}.$ So, when we write
$$ \frac{\dot{x}^2}{2} + \frac{1}{e} + x \log x = \mbox{constant} $$ we know that the constant is nonnegative. One may differentiate the equation to check it, using the original ODE. If the constant is $0,$ we have $\dot{x}=0, \; x = \frac{1}{e}.$ If the constant is positive, we have $$ x \log x \leq \mbox{constant} - \frac{1}{e}. $$
There is an oddity that happens because
$$ \lim_{x \rightarrow 0^+} x \log x = 0. $$
If we start with
$$ x(0) = \frac{1}{e}, \; \; \dot{x}(0) = -\sqrt{\frac{2}{e}}, $$
then $x$ continues to decrease forever but never quite reaches $0.$
If, Instead,
$$ x(0) = \frac{1}{e}, \; \; \dot{x}(0) < -\sqrt{\frac{2}{e}}, $$
then $x$ reaches $0$ in finite time and with $\dot{x}$ nonzero.