Step 1: Write down the function.
We are given $f(x) = \log\left(\dfrac{2x^2-3}{x} + \sqrt{\dfrac{4x^4-11x^2+9}{x^2}}\right)$. We need to check whether this is odd, even, a polynomial, or not a function.
Step 2: Simplify the expression inside the log.
Let $u(x) = \dfrac{2x^2-3}{x}$ and $v(x) = \sqrt{\dfrac{4x^4-11x^2+9}{x^2}}$. Note that $v(x) = \dfrac{\sqrt{4x^4-11x^2+9}}{|x|}$.
Step 3: Compute $f(-x)$.
Replace $x$ with $-x$: $u(-x) = \dfrac{2(-x)^2-3}{-x} = \dfrac{2x^2-3}{-x} = -u(x)$. Also, $v(-x) = \sqrt{\dfrac{4x^4-11x^2+9}{x^2}} = v(x)$ since $(-x)^2 = x^2$. So $f(-x) = \log(-u(x) + v(x))$.
Step 4: Add $f(x)$ and $f(-x)$.
Using the property $\log a + \log b = \log(ab)$: \[f(x) + f(-x) = \log\bigl[(u(x)+v(x))(-u(x)+v(x))\bigr] = \log\bigl[v(x)^2 - u(x)^2\bigr].\]
Step 5: Evaluate the product inside the log.
\[v(x)^2 - u(x)^2 = \frac{4x^4-11x^2+9}{x^2} - \frac{(2x^2-3)^2}{x^2}.\] Expand $(2x^2-3)^2 = 4x^4 - 12x^2 + 9$. So the numerator becomes $(4x^4-11x^2+9) - (4x^4-12x^2+9) = x^2$. Therefore $v(x)^2 - u(x)^2 = \dfrac{x^2}{x^2} = 1$.
Step 6: Conclude the nature of $f$.
Since $f(x) + f(-x) = \log(1) = 0$, we have $f(-x) = -f(x)$ for all $x$ in the domain. A function satisfying $f(-x) = -f(x)$ is called an odd function.
\[ \boxed{\text{Odd function}} \]