Functions create confusion

Replace $x$ by $\dfrac 1x$ . $3f(x)+2xf\left(\dfrac 1x\right)=\dfrac1{x^2}$ Now substitute value of $f\left(\dfrac 1x\right)$ from given eqn and rearrange for $f(x)$ to get:

$\Large f(x)=\dfrac{3-2x^5}{5x^2}$

$\therefore 100f(-2)=100\left(\dfrac{67}{20}\right)=\boxed{335}$

• $\large \displaystyle \text{Let } x = \frac{-1}{2}$

$\large \displaystyle 3 f(-2) + \frac{2 f \left(\frac{-1}{2} \right)}{\left(\frac{-1}{2} \right)} = \frac{1}{4}$

$\large \displaystyle \implies 3 f(-2) - 4 f \left(\frac{-1}{2} \right) = \frac{1}{4} \, .......... \, (1)$

• $\large \displaystyle \text{Let } x = -2$

$\large \displaystyle 3 f \left(\frac{-1}{2} \right) + \frac{2 f(-2)}{-2} = 4$

$\large \displaystyle \implies 3 f \left(\frac{-1}{2} \right) - f(-2) = 4 \, .......... \, (2)$

By Solving both the equations $(1)$ and $(2)$ for $f(-2)$ , We get

$\large \displaystyle f(-2) = \frac{67}{20} = \color{#D61F06}{A}$

$\huge \displaystyle \frac{67}{20} \times 100 = 67 \times 5 = \boxed{\color{#3D99F6}{3} \color{#20A900}{3} \color{#D61F06}{5}}$

Thanks @Hung Woei Neoh for correcting my mistake.

Let $x=-2$ , then:

$3f(-\frac{1}{2}) + 2\frac{f(-2)}{-2}=(-2)^2 \implies \boxed{3f(-\frac{1}{2}) - f(-2) = 4}$

Let $x=-\frac{1}{2}$ , then:

$3f(-2)+ 2\frac{f(-\frac{1}{2})}{-\frac{1}{2}} = (-\frac{1}{2})^2 \implies \boxed{3f(-2)-4f(-\frac{1}{2})=\frac{1}{4}}$

Solving the above pair of equations, we have $f(-2)=\frac{67}{20}$ . Now $\lfloor 100A \rfloor = \lfloor 100 \times \frac{67}{20} \rfloor = \lfloor 335 \rfloor = \boxed{335}$