6x^{2}+13x-5
------------ < 0
|5x^{3}-10|
I tried to do this with cases, but everything seems wrong, help would be greatly appreciated.
First of all, note that $\displaystyle \displaystyle \begin{align*} x \neq \sqrt[3]{2} \end{align*}$ (why?)
Then, since $\displaystyle \displaystyle \begin{align*} \left|5x^3 - 10\right| > 0 \end{align*}$ for all $\displaystyle \displaystyle \begin{align*} x \neq \sqrt[3]{2} \end{align*}$, we can multiply both sides of the inequality by it without changing the inequality sign.
$\displaystyle \displaystyle \begin{align*} \frac{6x^2 + 13x - 5}{\left|5x^3 - 10\right|} &< 0 \\ 6x^2 + 13x - 5 &< 0 \\ x^2 + \frac{13}{6}x - \frac{5}{6} &< 0 \\ x^2 + \frac{13}{6}x + \left(\frac{13}{12}\right)^2 - \left(\frac{13}{12}\right)^2 - \frac{5}{6} &< 0 \\ \left(x + \frac{13}{12}\right)^2 - \frac{169}{144} - \frac{120}{144} &< 0 \\ \left(x + \frac{13}{12}\right)^2 &< \frac{289}{144} \\ \left| x + \frac{13}{12}\right| &< \frac{17}{12} \\ -\frac{17}{12} < x + \frac{13}{12} &< \frac{17}{12} \\ -\frac{5}{2} < x &< \frac{1}{3} \end{align*}$
So the solution to your inequality is $\displaystyle \displaystyle \begin{align*} x \in \left( -\frac{5}{2} , \frac{1}{3} \right) \end{align*}$.