Suppose $f(x)$ is defined for all positive numbers $x$, and $2f(x-\frac{1}{x}) + f(\frac{1}{x}-x) = 3(x+\frac{1}{x})^2.$ Find $f(99)$.
Plugging in $x=1$, I was able to get $3f(0) = 3(1+1)^2 = 3\cdot 4$, so $f(0) = 4$. I've also tried letting $x-1/x = 99$ so that $2f(99) + f(-99) =3\cdot (99)^2$. However, I'm not sure how to proceed with these information (and I'm not sure what else to try).
If we sub in $-x$ for $x$, we get $2f\left ( \frac{1}{x}-x \right ) + f\left (x-\frac{1}{x} \right ) = 3\left ( -x-\frac{1}{x}\right )^2 = 3\left (x + \frac{1}{x} \right )^2$, so we now have two equations.
Note that $\left (x + \frac{1}{x} \right )^2 = \left (x - \frac{1}{x} \right )^2 + 4$. Thus, if we have $x-\frac{1}{x}=99$, we have the first equation as $2f(99)+f(-99) = 3(99^2+4)$, and second equation as $2f(-99)+f(99)=3(99^2+4)$.
Doubling the first equation and subtracting the second equation from it yields $3f(99) = 3(99^2+4)$, so $f(99)=99^2+4=9805$.