Let $f(x): \mathbb{R} \rightarrow \mathbb{R}$ be such that $g(x) =f(x)\times x^2$ is uniformly continuous. Prove that $\lim_{x\to\pm\infty}f(x)=0$.
Since $g$ is uniformly continuous than for every $\epsilon$ there must exist such $\delta$ that if $|x-y| < \delta$ than $|f(x)x^2-f(y)y^2|<\epsilon$. Applying $y=0$ and dividing by $x^2$ we conclude that as long as $|x|<\delta$ than $|f(x)|<\frac{\epsilon} {x^2}$ which seems almost like an end of a proof, but $x$ being bounded by $\delta$ opens it to a doubt.
Reproducing the approach in Anne Bauval's linked answer is fairly straightforward and "from the definition."
Given $\epsilon > 0$, let $\delta > 0$ be as you defined. (Without loss of generality let $\delta < 1$.) Let $\delta' = \delta/2$.
For $x > \delta'$, let $N_{x, \delta'} := \lfloor x/\delta' \rfloor$. Then we can partition the interval $[0, x]$ into $0 < \delta' < 2 \delta' < \cdots < N_{x, \delta'} \delta' < x$. Then,
\begin{align} |g(x)-g(0)| &= \left|g(x)-g(N_{x, \delta'} \delta') + \sum_{n=1}^{N_{x, \delta'}} (g(n\delta') - g((n-1)\delta'))\right| \\ &\le|g(x)-g(N_{x, \delta'} \delta')| + \sum_{n=1}^{N_{x, \delta'}} |g(n\delta') - g((n-1)\delta')| \\ &\le (N_{x, \delta'} + 1) \epsilon \\ &\le \frac{4x\epsilon}{\delta}. \end{align} Plugging in the definition of $g$ yields \begin{align} x^2 |f(x)| &\le \frac{4x\epsilon}{\delta} \\ |f(x)| &\le \frac{4\epsilon}{\delta x} \end{align}
This bound holds for all $x > \delta'$ so $\lim_{x \to \infty} f(x) = 0$.