Answer
$-\sqrt{2} < x < \sqrt{2}$
Work Step by Step
Factor out $e^{x}:$
$e^{x}(x^{2}-2) < 0$
The function $f(x)=e^{x}$ is always positive, never zero,
so LHS $ < 0$ only if
$x^{2}-2 < 0$
$x^{2}<2$
This is valid only if $|x| < 2$, that is, if
$-\sqrt{2} < x < \sqrt {2}$