I would like to check if my proof is OK and whether there was a shorter way to prove this.
This is my attempt:
First consider $I=(0,\infty)$.
Notice, $f(x)$ is derivable as a product/sum/and composition of derivable (elementary) functions in $I$.
Notice, $f^{'}(x) = 1+\ln(x)+3x\cos(x^3)-{\sin(x^3) \over x^2}$. It's easy to see $\lim_{x \to \infty}f^{'}(x) = \infty$.
Assume with contradiction $f(x)$ is uniformly continuous in $I$. Let $\epsilon \gt 0 \implies \exists \delta \gt 0$ s.t $\forall x_1,x_2 \in I$:
if $|x_2 - x_1| \lt \delta \implies |f(x_2)-f(x_1)| \lt \epsilon$.
From $\exists \lim_{x \to \infty}f^{'}(x)$ then $\forall M \gt 0 \in I$ exists some $x_R$ s.t if $x \gt x_R$ then $f^{'}(x) \gt M$.
Take $M = {\delta \over 2}$. Let $x_2 \gt x_1 \gt x_R \in I$ s.t $x_2-x_1 = {\delta \over 2}$.
Consider $[x_1,x_2]$. From MVT we have ${f(x_2)-f(x_1) \over x_2-x_1} = {|f(x_2)-f(x_1)| \over |x_2-x_1|} = f^{'}(c)$ for some $c \in (x_1,x_2)$.
Therefore we have: $|f(x_2)-f(x_1)| \gt {\delta \over 2}\times{2 \over \delta} = 1$.
This is obviously a contradiction.
Now consider $I=(0,20)$.
$f(x)$ is continuous as a product/sum and composition of continuous functions in $I$.
Therefore $f(x)$ is uniformly continuous $\iff \exists \lim_{x \to 0^+}f(x)$ and $\exists \lim_{x \to 20^-}f(x)$.
$\lim_{x \to 0^+}f(x) = \lim_{x \to 0^+}x\ln(x) + \lim_{x \to 0^+}{\sin(x^3) \over x} = \lim_{x \to 0^+}{\ln(x) \over {1 \over x}} + \lim_{x \to 0^+}{\sin(x^3) \over x^3}x^2 = -x+x^2 = 0$.
$\lim_{x \to 20^-}f(x) = 20\ln(20) - {1 \over 20}\sin(20^3)$.
Therefore $\exists \lim_{x \to 0^+}f(x)$ and $\exists \lim_{x \to 20^-}f(x) \implies f(x)$ is uniformly continuous in $I$.
The limit of ${1\over x}\sin(x^3)$ is equal $0$ at $0$ and the limit at $\infty$ is equal $0.$ Therefore ${1\over x}\sin(x^3)$ is uniformly continuous on $(0,\infty).$ It suffices to study $g(x)=x\log x.$ The limit of $g(x)$ at $0$ is equal $0.$ Therefore the function $g(x)$ can be extended to a continuous function on $[0,\infty).$ Hence it is uniformly continuous on every interval $[0,N].$ Concerning $[0,\infty)$ let $x_n=n$ and $y_n=n+a_n.$ Then $$g(y_n)-g(x_n)=n[\log(n+a_n)-\log n]+a_n\log(n+a_n)\\ = n\log(1+n^{-1}a_n)+a_n[\log n+\log(1+n^{-1}a_n)]\\ \sim a_n+a_n\log n+n^{-1}$$ For $a_n=(\log n)^{-1}$ the limit of the last expression is equal $1.$ Hence the function $g$ is not uniformly continuous.