Let $R$ denote a non-negative rational number. Determine a fixed set of integers $a,b,c,d,e,f$, such that for every choice of $R$,
$\left|\frac{aR^2+bR+c}{dR^2+eR+f}-\sqrt[3]{2}\right|<|R-\sqrt[3]{2}|$
The given inequality seemed too much similar to limits so I applied R tending to $\sqrt[3]{2}$. And then I tried solving the equation I get however I am unable to solve for integers. Any help please
(Too long for a comment.) The question is missing a lot of context, so it is hard to guess if the following even applies at all. But FWIW the given form reminisces of an iterated approximation steadily approaching $\,\sqrt[3]{2}\,$.
Newton-Raphson would come to mind first, but the canonical $\,f(x) = x^3-2\,$ doesn't give $\,x - f(x) / f'(x)\,$ in the prescribed form of a ratio of two quadratics.
Another way to "guess" an answer could be to start from the identity $\,a^3-b^3=(a-b)(a^2+ab+b^2)\,$, so $\,\sqrt[3]{2}=1+1 \big/ \left(\left(\sqrt[3]{2}\right)^2+\sqrt[3]{2}+1\right)\,$ i.e. $\,\sqrt[3]{2}$ is a fixed point of $\,f(x)=(x^2+x+2)/(x^2+x+1)\,$ which corresponds to $\,a=b=d=e=f=1, c=2\,$ and does indeed appear to converge as required. Formally proving that convergence would take some more work, though.
[ EDIT ] Following up on OP's insight that...
Formalizing the above, let $\,f(x)=\dfrac{ax^2+bx+c}{dx^2+ex+f}\,$ and consider the sequence defined by $\,r_{n+1}=f(r_n)\,$ for some $\,r_0 \in \mathbb{R}\,$. Furthermore, suppose that $\,r_n \to r = \sqrt[3]{2}\,$, then passing the recurrence relation to the limit gives:
$$r=f(r) \;\;\iff\;\; dr^3+(e-a)r^2+(f-b)r-c = 0$$
But the minimal polynomial of $\,r=\sqrt[3]{2}\,$ over $\,\Bbb Z[X]\,$ is $x^3 - 2$, so it follows that $\,\begin{align}\begin{cases}2d &= c \\ e&=a \\f&=b\end{cases}\end{align}\,$ and therefore $\,f(x) = \dfrac{ax^2+bx+2d}{dx^2+ax+b}\,$ or, choosing WLOG $\,d=1\,$, $\,f(x) = \dfrac{ax^2+bx+2}{x^2+ax+b}\,$ .
The case given in the original answer $\,f(x)=\dfrac{x^2+x+2}{x^2+x+1}\,$ corresponds to $\,a=b=1\,$ above.