Does $xe^{1/\log(x)}$ have an oblique asymptote when x tends to infinity? Does $xe^{1/\log(x)}$ have an oblique asymptote when $x$ tends to infinity?
 If so, what is the equation of this asymptote and how can we find it?
If $f(x) = xe^{1/\ln x}$ has an oblique asymptote $y=Ax+B$, then $A = \lim_{x\to\infty} f(x)/x$, but the existence of this limit is not enough to conclude that there is an oblique asymptote.

We also need that $B = \lim_{x\to\infty} f(x)-Ax$ exists.

In your example, we have
$$A = \lim_{x\to\infty} \frac{xe^{1/\ln x}}{x} = \lim_{x\to\infty} e^{1/\ln x} = 1$$
but
\begin{align}
B &amp;= \lim_{x\to\infty} xe^{1/\ln x} - x \\
  &amp;= \lim_{x\to\infty} x(e^{1/\ln x}-1) \qquad [t=1/\ln x] \\
  &amp;= \lim_{t\to 0^+} e^{1/t}(e^t-1) = +\infty,
\end{align}
so your $f$ doesn't have an oblique asymptote when $x\to\infty$.
Using Ruslan's definition,
we want to see if
$\lim_{x \to \infty} \dfrac{x e^{1/\ln x}}{x}
$
exists.

$\dfrac{x e^{1/\ln x}}{x}
=e^{1/\ln x}
$
and,
since 
$\lim_{x \to \infty} 1/\ln x
= 0
$,
$\lim_{x \to \infty} e^{1/\ln x}
= e^0
=1
$.

So it does have a 
oblique asymptote
with slope $1$.