Existence of Partials Imply the Existence of Gradient Vector? Let $f$ be a scalar function of three variables. Then the gradient vector is defined by:

I read here that the existence of partial derivatives at some point $(x_0, y_0, z_0)$ does not imply the existence of the gradient vector at $(x_0, y_0, z_0)$. How is this possible, since the gradient vector is just a specific type of sum of the partial derivatives?

The correct definition of the gradient vector is the following.

Let $f: X \subset \mathbb{R}^n \to \mathbb{R}$ be a function that is differentiable at point $\mathbf{x}_0 \in \operatorname{int} X$. Then gradient of $f$ at point $\mathbf{x}_0$ is the following vector
$$\nabla f(\mathbf{x}_0) = \sum_{i=1}^n \mathbf{e}_i \frac{\partial f}{\partial \mathbf{e}_i}(\mathbf{x}_0) = \sum_{i=1}^n \mathbf{e}_i \frac{\partial f}{\partial x_i}(\mathbf{x}_0) = \begin{bmatrix} \frac{\partial f}{\partial x_1}(\mathbf{x}_0) \\ \frac{\partial f}{\partial x_2}(\mathbf{x}_0) \\ \vdots \\ \frac{\partial f}{\partial x_n}(\mathbf{x}_0) \end{bmatrix},$$
where $\{\mathbf{e}_i\}_{i=1}^n$ is some orthonormal basis in $\mathbb{R}^n$.

We do need the requirement of differentiability of $f$ at point $\mathbf{x}_0$ because it is easy to show that only in this case the expression above will be invariant when passing from one orthonormal basis to another (hence, it will be correctly defined vector object in $\mathbb{R}^n$). I mean that this requirement guarantees that if we change one orthonormal basis $\{\mathbf{e}_i\}_{i=1}^n$ in $\mathbb{R}^n$ to another orthonormal basis $\{\mathbf{h}_i\}_{i=1}^n$ in $\mathbb{R}^n$, then the expression for the gradient vector will change from $\sum_{i=1}^n \mathbf{e}_i \frac{\partial f}{\partial \mathbf{e}_i}(\mathbf{x}_0)$ to $\sum_{i=1}^n \mathbf{h}_i \frac{\partial f}{\partial \mathbf{h}_i}(\mathbf{x}_0)$ , such that these two expressions will be exactly equal: $\sum_{i=1}^n \mathbf{e}_i \frac{\partial f}{\partial \mathbf{e}_i}(\mathbf{x}_0) = \sum_{i=1}^n \mathbf{h}_i \frac{\partial f}{\partial \mathbf{h}_i}(\mathbf{x}_0)$, i.e. they describe the same object in different basises.
If all partial derivatives exist at point $\mathbf{x}_0$, but $f$ is not differentiable at this point, then there will no be such invariance property and the set $\left(\frac{\partial f}{\partial x_1}(\mathbf{x}_0), \ldots, \frac{\partial f}{\partial x_n}(\mathbf{x}_0)\right)^\mathrm{T}$ technically will not be a correctly defined vector object at all.
