I do not know what the definition of a critical value is.
BUT
When x=0
f(0)= 0^(1/9) - 0^(-8/9)
=01/9−108/9=0−undefined=undefined
So there is probably an asymptote at x=0
f(x)= x^(1/9) - x^(-8/9)
f(x)=x1/9−x−8/9f′(x)=x−8/99+8x−17/99f′(x)=19x8/9+89x17/9=0
By inspection I do not think this is ever the case.
So there are no stationary points.
Here is the graph.
