We need to solve the inequality $\frac{3}{x} + \frac{x}{3} > \frac{3(x+3)^2 - x(9+2x^2)}{3x^2+9x}$.

1. Problem Description

We need to solve the inequality $\frac{3}{x} + \frac{x}{3} > \frac{3(x+3)^2 - x(9+2x^2)}{3x^2+9x}$.

2. Solution Steps

First, we simplify the right-hand side of the inequality.

The numerator is $3(x+3)^2 - x(9+2x^2) = 3(x^2+6x+9) - 9x - 2x^3 = 3x^2 + 18x + 27 - 9x - 2x^3 = -2x^3 + 3x^2 + 9x + 27$.

The denominator is $3x^2+9x = 3x(x+3)$.

So the right-hand side is $\frac{-2x^3+3x^2+9x+27}{3x(x+3)}$.

Now, we rewrite the inequality as

$\frac{3}{x} + \frac{x}{3} > \frac{-2x^3+3x^2+9x+27}{3x(x+3)}$.

The left-hand side is $\frac{9+x^2}{3x}$. Thus, we have

$\frac{9+x^2}{3x} > \frac{-2x^3+3x^2+9x+27}{3x(x+3)}$.

Multiplying both sides by $3x(x+3)$ gives

$(9+x^2)(x+3) > -2x^3+3x^2+9x+27$ if $x(x+3)>0$, i.e., $x<-3$ or $x>0$.

$(9+x^2)(x+3) < -2x^3+3x^2+9x+27$ if $x(x+3)<0$, i.e., $-3<x<0$.

Expanding the left side gives

$9x + 27 + x^3 + 3x^2 > -2x^3+3x^2+9x+27$.

So $x^3 + 9x + 27 + 3x^2 > -2x^3+3x^2+9x+27$.

Then $x^3 > -2x^3$, which implies $3x^3 > 0$, so $x^3 > 0$, and thus $x>0$.

If $x<-3$ or $x>0$, then $(9+x^2)(x+3) > -2x^3+3x^2+9x+27$ becomes $x>0$.

Since $x>0$ is contained in $x<-3$ or $x>0$, $x>0$ is part of the solution.

If $-3<x<0$, then $(9+x^2)(x+3) < -2x^3+3x^2+9x+27$ becomes $x>0$.

However, this contradicts $-3<x<0$, so there is no solution in this case.

Therefore, the solution is $x>0$.

Also, $x$ cannot be $-3$ or $0$.

3. Final Answer

$x>0$