How to find the oblique asymptote of root of a function?

Yes. If $f$ has an oblique asymptote (call it $y=ax+b$), you will have: $$a=\lim_{x\to\pm\infty}\frac{f(x)}{x}$$

$$b=\lim_{x\to\pm\infty} f(x)-ax$$

In your example, $\displaystyle\lim_{x\to+\infty}\frac{\sqrt{4x^2+x+6}}{x}=2$ and $\displaystyle\lim_{x\to+\infty}\sqrt{4x^2+x+6}-2x=\frac{1}{4}$

The asymptote as $x\to+\infty$ is therefore $y=2x+\dfrac{1}{4}$

The answer of @Julien is perfect, but here’s another outlook. Take your function, and factor out $4x^2$ from the radicand, getting $2x\sqrt{1+1/(4x) + 3/(2x^2)}=2x(1+\frac{1}{4}x^{-1}+\frac{3}{2}x^{-2})^{1/2}$. For the (positive) asymptote, you’re interested in cases where $x^{-1}$ is tiny, so you can approximate the radical very well with the Taylor expansion $(1+A)^{1/2}=1+\frac{1}{2}A-\frac{1}{8}A^2+\cdots$. Setting $A=\frac{1}{4}x^{-1}+\frac{3}{2}x^{-2}$ and looking only at the constant and the $x^{-1}$-term, you get $2x(1+\frac{1}{8}x^{-1}+\cdots)$, the same result that @Julien announced.