Given, $y=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}$
$y$ is defined for
$\log _{10}(1-x) \neq 0$ and $x+2 \geq 0$
$\Rightarrow \, 1-x \neq 1 x \geq-2 $
$x \neq 0$
and $1-x>0$ and $1-x \neq 0$
$x<1$ and $x \neq 1$
$\therefore y$ is defined for values $x$ for which both the terms of the function is defined
$\therefore $ y is defined, when
$-2 \leq x<\,0$
And $ 0<\,x<\,1$