Answer
$(-\frac{8}{7},0)$
Work Step by Step
1. $f(x)=\frac{2\sqrt {x+2}}{3\sqrt[3] x}+\frac{\sqrt[3] {x^2}}{2\sqrt {x+2}}\lt0$
$f(x)=\frac{4(x+2)+3x}{6\sqrt[3] x\sqrt {x+2}}=\frac{7x+8}{6\sqrt[3] x\sqrt {x+2}}\lt0$
2. The denominator $\sqrt {x+2}$ requires that $x>-2$, cut points $-8/7(-1.1), 0$ make a table as shown.
3. Solution $(-\frac{8}{7},0)$
