Answer
$(-∞, -1 - \sqrt 3) U [0, -1 + \sqrt 3) $
Work Step by Step
$\frac{x}{x^2 + 2x - 2} \leq 0$
Find the zeros of the expressions in the numerator AND the denominator
$x^2 + 2x + 1 = 3$
$(x+1)^2 = 3$
$(x+ 1) = \sqrt 3$
$x = -1 + \sqrt 3, -1 - \sqrt 3$
$x^2 + 2x -2$ = $(x+ 1 -\sqrt 3) (x+ 1 + \sqrt 3)$
Test numbers in between those zero values to determine if the function is negative or positive
$(-∞, -1 - \sqrt 3)$ $\frac{(-)}{(-)(-)} = (-)$
$(-1 - \sqrt 3, 0]$ $\frac{(-)}{(-)(+)} = (+)$
$[0, -1 + \sqrt 3]$ $\frac{(+)}{(-)(+)} = (-)$
$(-1 + \sqrt 3, ∞)$ $\frac{(+)}{(+)(+)} = (+)$
Thus the solution is $(-∞, -1 - \sqrt 3) U [0, -1 + \sqrt 3) $
