Calculating limit $\lim\limits_{x\to\infty}\frac{3x^2-\frac{3}{x^2+1}-4f'(x)}{f(x)}$ for an unknown function.

Since $f''(x)=2~~for ~~x>0$, Therefore $f$ has the form $f(x)=x^2+bx +c~~~for~~~x>0$

Since the limit is at $+\infty$ it suffices to consider $x>0$ $$\lim_{x\to\infty}\frac{3x^2-\frac{3}{x^2+1}-4f'(x)}{f(x)} = \lim_{x\to\infty}\frac{3x^2-\frac{3}{x^2+1}-8x-4b}{x^2+bx +c} =3$$

From $f''(x)=2$, we know that $f$ is a quadratic function with leading term $x^2$. There will be no cancellation of $3x^2$ at the numerator, and by ignoring low order terms, the expression simplifies to

$$\frac{3x^2}{x^2}.$$

The two given conditions mean $\;\lim\limits_{x\to\infty}f(x)=\infty\;$ (why?), and then, from your work, we reached

$$\lim_{x\to\infty}\frac{6+\frac{6(x^2+1)^2-6x×2(x^2+1)×2x}{(x^2+1)^4}+0}{f''(x)}=\frac{6+0+0}2=3$$

I don't really understand what you did after the above part in your question...but the middle summand in the numerator above is

$$\frac{6(x^2+1)-24x^2}{(x^2+1)^3}\xrightarrow[x\to\infty]{}0$$