Interpreting the problem exactly as written:
We immediately know there's not going to be any real solutions to the problem because any (real number) squared must be positive, and 9 times a positive real number is still positive, so it cannot be the case that \(9x^2 < 0\) unless \(x\) is complex. Therefore, we can guarantee that \(x \ne 0\), such that dividing all sides of the inequality by \(x^2\) is a legal move. If we do that, we get \(x^2 < 9 < 0\) which is impossible.
On the other hand, your workings seem to indicate that the problem was actually:
\(\displaystyle x^4 \: {\color{red}{\mathbf{-}}} \: 9x^2 < 0\)
If that's the case, then everything you've done is good. Let's now temporarily suppose it was in equality and find the roots of:
\(\displaystyle x^2(x - 3) (x + 3) = 0\)
It should be clear that the roots are 0, 3, and -3. Next we can use this information to look at cases. What if \(x < -3\)? We'd have:
\(\displaystyle x^2 > 0 \\ x - 3 < 0 \\ x + 3 < 0 \\~\\
\text{{Positive}} \cdot \text{{Negative}} \cdot \text{{Negative}} = \text{{Positive}}\)
Thus, we can conclude that \(x^2(x-3)(x+3) > 0\) in this interval. Now let's consider the next interval. What if \(x\) is between the roots? That is, what if \(-3 < x < 0\)? I'll leave this and the rest of the intervals to you to work out, but see what you find when all is said and done.