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Let $f(x)=x+log_e(x)-xlog_e(x)$
I am confused if I want to know whether $lim_{xrightarrow infty }f(x)$ is $-infty$ or $infty$

I used the following concept $f(x)=x+log_e(x)-xlog_e(x)$

$lim_{xrightarrow infty }f(x)$=$infty+log_e(infty)(1-infty)$.

After this step I am confused

If you use extended reals https://en.wikipedia.org/wiki/Extended_real_number_line#Arithmetic_operations then the expression $infty+log_e(infty)(1-infty) = infty + infty - infty times infty = infty - infty$ which is undefined. So you can't get an answer like this.

However, for any finite $x > 0 $ we have an inequality $ log_e(x) < x$ so that
$f(x)=x+log_e(x)-xlog_e(x) < x+ x-xlog_e(x) = x(2 - log_e(x))$

Now when you let $x to infty $ you have $ x(2 - log_e(x)) = infty(-infty) = -infty$

So, $lim_{xrightarrow infty }f(x) = - infty$

At an introductory level it is better to keep the discourse within $Bbb R$ and treat $pm infty$ as figures of speech, or abbreviations. For example, $lim_{xto infty}f(x)=infty$ is an abbreviation for $$forall rin Bbb R,exists sin Bbb R,forall xin Bbb R,(x>simplies (f(x)>r).$$

We cannot do arithmetic on figures of speech. We have to be able to translate them into their complete and precise meanings about actual mathematical objects.

If $x>e^2$ then $(ln x>2land x-1>0)$ so $(ln x)(x-1)>2(x-1)$.

So if $x>e^2$ then $$x+ln x-xln x=x-(ln x)(x-1)<x-2(x-1)=$$ $$=2-x.$$

Now, given $rin Bbb R,$ let $s=max (-r+2,e^2).$ If $x>s$ then $$x+ln x-xln x<$$ $$<2-x quad text {(because}, x>e^2)$$ $$<r quad text {(because} , x>-r+2).$$

Therefore $$forall rin Bbb R,exists sin Bbb R, forall xin Bbb R,...$$ $$...(x>simplies x+ln x-xln x<r).$$ Which we abbreviate as $lim_{xto infty}(x+ln x-xln x)=-infty.$

$$f(x) = x + log x - x log x = 1 + (-1 + x + log x - x log x) = 1 - (x - 1)(log x - 1).$$ For any $N > 0$, $x > N+1$ implies $x-1 > N$, and $x > e^{N+1}$ implies $log x - 1 > N$. Consequently, given some number $N > 0$, the choice $x > max(N+1, e^{N+1})$ guarantees $f(x) < 1 - N^2$, which of course implies $f$ decreases without bound as $x$ increases without bound.

Note the only property of $log$ we have relied upon here is that it is an increasing function that is unbounded above.