Prove that $f: \mathbb{R} \to \mathbb{R}, f(x) = x^\frac{1}{9}$ is not a differentiable function Prove that the following function is not a differentiable function: $$f: \mathbb{R} \to \mathbb{R}, f(x) = x^\frac{1}{9}$$
I believe all I have to show is one point in the domain where the function is not differentiable:
Hence I have the following proof:
$$ f'(x) =\frac{1}{9x^\frac{8}{9}} $$
At $ x = 0$, $f'(x)$ is undefined.
Hence, $f(x)$ is not a differentiable function.
Is this enough or do I need to show that the $\lim\limits_{h\to0} \frac{f(x_0 + h) - x_0}{h}$ does not exist in some other way.

Yes, you must show that the limit $\lim_{x\to 0}\frac{f(x)-f(0)}{x}$does not exist. 
Note that $f'(x)$ exists for all $x\ne 0$ however at $x=0$, differentiability is doubtful that is to say it is apriori not known whether $f'(0)$ exists or not and if it does exist then what form it will be in. 
What you have shown above is that $f'(x) =\frac{1}{9x^\frac{8}{9}}$ does not exist at $x=0$, which has nothing to do with $f'(0)$ as we don't even know apriori about existence of $f'(0)$ or the kind of form $f'(0)$ will be in if it exists.
Therefore, by definition of differentiability at $x=0$, $f'(0)$ exists if and only if $\lim_{x\to 0}\frac{f(x)-f(0)}{x}$ exists. 
Clearly, $\frac{f(x)-f(0)}{x}=\frac{x^{1/9}-0}{x}=\frac 1{x^\frac 89}$ (Note that it is different from $\frac{1}{9x^\frac{8}{9}}$) and therefore the limit does not exist finitely, which is to say that $f'(0)$ is non-existent.
