Draw the graph of `y=x+(1)/(x)`.

Explanation

We have `y=f(x)=log_(e)(sqrt(1-x^(2)-x))`
`f(x)` is defined if `sqrt(1-x^(2)) - x gt 0` and `1-x^(2) ge0`
For `1-x^(2) ge0 , -1 le x le1`
`sqrt(1-x^(2))-x gt 0`
For (0,1),
`1-x^(2)gt x^(2)`
or `0 lt x lt 1/sqrt(2)`
Thus, the domain of the function is `[-1, 1/sqrt(2)]`
`f^(')(x) = ((-x/sqrt(1-x^(2)))-1)/(sqrt(1-x^(2))-x)`
`=(-x-sqrt(1-x^(2)))/((sqrt(1-x^(2))sqrt(1-x^(2))-x)`
`f^(')(x)=0 therefore -x=sqrt(1-x^(2)`
`therefore x=-1/sqrt(2)`
`f^(')(x) gt 0` for `x in (-1,-1/sqrt(2))`
`f^(')(x) lt 0` for `x in (-1/sqrt(2), 1/sqrt(2))`
So `x=-1/sqrt(2)` is the point of maxima.
`f(-1) = f(0)=0`
`underset(x to 1/sqrt(2))"lim"log_(e)(sqrt(1-x^(2))-x) =- infty (therefore underset(x to 1/sqrt(2))"lim"(sqrt(1-x^(2))-x)=0)`
Thus, `x=1/sqrt(2)` is an asymptote.
From the above discussion, the graph of the function is as shows in the following figure.