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I've been slowly losing my mind on how to do this question, any help would be greatly appreciated.

Solve the inequality $$frac{2}{x} > 3 x.$$

To solve the equation
$$
frac{2}{x} > 3x
$$
we could multiply be $x$ to obtain
$$
2 > 3x^2.
$$
But you have to be careful!
If $x < 0$, the inequality is reversed.
In other words:
$$
frac{2}{x} > 3x
iff begin{cases}
2 < 3x^2 & x < 0 \
2 > 3x^2 & x > 0.
end{cases}
$$
Notice the equation isn't defined for $x = 0$, because you can't divide by $0$.

Case 1: $x > 0$.
Then we have
$$
2> 3x^2
iff x^2 < frac{2}{3}
iff x in left(- sqrt{ frac{2}{3}}, sqrt{ frac{2}{3}}right).
$$
But, since $x > 0$, our first solution interval is $I_1 := left(0, sqrt{ frac{2}{3}}right)$.

Case 2: $x < 0$.
Then we have
$$
2 < 3x^2
iff x^2 > frac{3}{2}
iff x > sqrt{frac{3}{2}} quad text{and} qquad x < -sqrt{frac{3}{2}}
$$
But again, since $x < 0$, our second (and final) solution interval is $I_2 := left(- infty, -sqrt{frac{3}{2}}right)$.

So our solution is $I_1 cup I_2 =left(- infty, -sqrt{frac{3}{2}}right) cup left(0, sqrt{ frac{2}{3}}right)$.

We observe that necessarily $xneq 0$ whence we can rewrite the inequality as
$$
frac{2}{x}-3x>0ifffrac{2x-3x^3}{x^2}>0iff 2x-3x^3>0iff x(2-3x^2)>0
$$
which you can solve.

Usually you're supposed to give some context and explain what you've tried so far. However, I will take your word that you have been "slowly losing" your mind and give an answer.

Since this is a nonlinear inequality, there is no "cookie cutter" way to solve it. But you can get an idea of what to do by graphing $y_1=2/x$ and $y_2=3x$, and observing where $y_1>y_2$.

See here for a graph with $y_1=2/x$ in red and $y_2=3x$ in blue. Where does the red curve lie above the blue curve?

Well, you can see that there are two intersection points. But the red lies above the blue from $-infty$ up to the first intersection point, and then again from zero to the second intersection point.

To find the intersection points, set $2/x=3x$ and solve.

Can you write down the final answer?