Graphing error found 
$f(x)=e^x|\ln(x)|$ is defined over $(0,+\infty)$.
How is there a negative part of the function shown by graphing it?!
 A: You're right that
$$f(x) = e^x|\log(x)|$$
is only defined on $(0,+\infty$), so the "problem" lies with the graphing software.

Below is a plausible explanation that seems to match the graph drawn. Many calculators and software follow the steps listed elow, e.g. my TI-83, WolframAlpha and GeoGebra.
Step 1: interpreting the logarithm of a negative real number.
The logarithm on the reals is not defined when $x \leq 0$, but one does have the principal value of the complex logarithm:
$$ \operatorname{Log}: \mathbb{C}\setminus \{0\} \to \mathbb{C}: z = re^{i\theta} \mapsto \log(r) + i\theta,$$
where $r > 0$ and $\theta \in (-\pi,\pi]$.
In particular, for a real $x < 0$ you have
$$ \operatorname{Log}(x) = \operatorname{Log}(|x|e^{\pi i}) = \log(|x|) + i\pi.$$
Calculators and software that can calculate this complex logarithm, will often default to using this when asked to calculate the logarithm of a negative number.
Step 2: interpreting the absolute value.
While the absolute value $|\;.\,|$ is defined on the real numbers, there is natural extension to the complex numbers: the modulus function
$$|\;.\,|: \mathbb{C} \to \mathbb{R}^+: z = a+bi \mapsto \sqrt{a^2+b^2}.$$
Again, many calculators and software will default to this function when asked to calculate the absolute value of a nonreal number.
In the case of $\operatorname{Log}(x)$ with $x < 0$ a real number, we have:
$$|\operatorname{Log}(x)| = |\log(|x|) + i\pi| = \sqrt{\log(|x|)^2 + \pi^2}.$$
Result:
The function $$f: (0,+\infty) \to \mathbb{R}: x \mapsto e^x |\log (x)|$$
gets extended to the function
$$f: \mathbb{R}\setminus \{0\} \to \mathbb{R}: x \mapsto \left\{\begin{array}
{ll}
e^x |\log (x)| &\text{ if } x > 0\\
e^x \sqrt{\log(|x|)^2 + \pi^2} &\text{ if } x < 0
\end{array}\right.,$$
which is what you get on your graphing app.
