How can I show that there is no Lipschitz gradient $\nabla f$ for any $C < \infty$ for the function $ f: \mathbb{R}^2 \rightarrow \mathbb{R}$
$$f(\text{x}) = {\Vert\text{x}\Vert}^{3/2}$$
Updated with gradient calculation below:
$$f(\text{x}) = ((\sum_{k=1}^2 x_k^2)^{1/2})^{3/2} =(\sum_{k=1}^2 x_k^2)^{3/4}$$
$$\begin{aligned} \nabla_j f(\text{x}) &= \partial/\partial x_j(\sum_{k=1}^2 x_k^2)^{3/4}\\ &= \frac{3}{4} (\sum_{k=1}^2 x_k^2)^{-1/4} (2x_j)\\ &=\frac{3}{2} x_j (\sum_{k=1}^2 x_k^2)^{-1/4}=\frac{3}{2}\frac{x_j}{\sqrt{\Vert x \Vert}} \end{aligned}$$
Updated to add definition of Lipschitz Gradient for a continuously differentiable function $f$, where $ L \in [0, \infty)$
$$ \Vert\nabla f(\text{x}) - \nabla f(\text{y})\Vert \leq L \Vert \text{x} - \text{y} \Vert$$
So you have
$$\nabla_j f(x)=\frac{3}{2}\frac{x_j}{\sqrt{\Vert x \Vert}}.$$
Taking $x=(t,0)$ with $t \gt 0$ you get
$$\nabla f(x)=\frac{3}{2}(\sqrt t, 0)$$ and the desired result as $t \mapsto \sqrt t$ is not Lipschitz. Last statement is a consequence of $\lim\limits_{t \to 0^+} \frac{\sqrt t }{t} = \infty$.