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.

As others pointed out, the non-existence of $\lim_{x\to 0}f'(x)$ does not imply that $f$ is not differentiable at $x=0$. However, the following is correct and can be applied here:

Let $f: (a-h, a+h) \to \Bbb R$ be continuous, and differentiable on $(a-h, a+h) \setminus \{ a \}$. If $\lim_{x\to a}|f'(x)| = \infty$ then $f$ is not differentiable at $x=a$.

Proof: This is a consequence of the mean-value theorem: Assume that $f'(a)$ exists, and let $x_n$ be a sequence in  $(a-h, a+h) \setminus \{ a \}$ which converges to $a$. Then
$$
 \frac{f(x_n)-f(a)}{x_n-a} = f'(c_n)
$$
for some $c_n$ between $a$ and $x_n$. The left-hand side converges to $f'(a)$, whereas the right-hand side is unbounded for $n \to \infty$. This is a contradiction.

In your case, $f(x) = x^{1/9}$ is continuous on $\Bbb R$, differentiable on $\Bbb R \setminus \{ 0\}$, and
$$
 \lim_{x\to 0}|f'(x)| = \lim_{x\to 0}\frac{1}{9|x|^{8/9}} = \infty \, ,
$$
which implies that $f$ is not differentiable at $x=0$.
