Answer
The domain of definition is: $(-2, 1]$
Work Step by Step
$\sqrt[4] {\frac{1-x}{2+x}}$
Our restriction here is, that the whole expression under $4^{th}$ root must be non-negative and denominator must not be $0$.
$\left\{ \begin{array}{ll} \frac{1-x}{2+x} \geq0\\ 2+x\ne0 \end{array} \right. $
For the first restriction to be satisfied we have $2$ possible intervals:
$\left\{ \begin{array}{ll} 1-x <0\\ 2+x<0 \end{array} \right. $ $\left\{ \begin{array}{ll} x >1\\ x0\\ 2+x>0 \end{array} \right. $ $\left\{ \begin{array}{ll} x <1\\ x>-2 \end{array} \right. $
In this case, we have domain $[-2, 1]$
But note, our $2^{nd}$ restriction doesn't permit us to include $-2$ in the interval:
$2+x\ne0$
$x\ne-2$
So, at last the domain of definition is: $(-2, 1]$
