Recently reading a set of lecture notes on vector calculus, which is a topic I am already familiar with. However during this I came across this representation of the gradient vector...
$$\frac{\partial \phi}{\partial x^{i}} {e}_{i} = \nabla \phi $$
I don't understand why on the suffix for the derivative it is raised however on the basis vector it is lowered. I have a brief grasp of the metric tensor having studied special relativity, however I don't feel like it would play any role here.
On
I will assume we are working in $\Bbb R^3$, just to keep things brief, but this generalizes very easily to higher dimensions, and with some work to general smooth manifolds.
Say we are given a point $x$ in a 3-dimensional real vector space. Cool.
Now we impose a coordinate system on this vector space. What does that mean? It means we pick 3 linearly independent vectors $e_1, e_2, e_3$ to be our basis vectors, and we use those, together with coefficients, to express any other vector. For instance, we have $$ x = x^1e_1 + x^2e_2 + x^3e_3 = x^ie_i $$ Why is one index upper and one lower? Because they ultimately always belong in pairs like this, with one upper and one lower, and having them all be lower or all be upper would make it difficult to keep track. Why are the coefficients upper and the vectors lower? Maybe someone had a good reason back in the day, but ultimately it doesn't matter: we have to choose one way or the other, this is the way that has been chosen.
Now we apply a function $\phi:\Bbb R^3\to \Bbb R$ to our vector space. This function has a gradient. The standard way to calculate this gradient is to first express $\phi$ as a function of the three separate coordinates rather than as a function of points in 3-dimensional space. Then we can differentiate in each coordinate direction and use the resulting numbers as coefficients in a new vector.
Expressing $\phi$ as a function of the three separate coordinates takes the form $\phi(x^1, x^2, x^3)$. And then we differentiate this with respect to each component to get the three new coefficients $\frac{\partial \phi}{\partial x^1}, \frac{\partial \phi}{\partial x^2}, \frac{\partial \phi}{\partial x^3}$. Using those as the coefficients of a new vector gives us $$ \nabla \phi = \frac{\partial \phi}{\partial x^1}e_1 + \frac{\partial \phi}{\partial x^2}e_2 + \frac{\partial \phi}{\partial x^3} e_3 = \frac{\partial \phi}{\partial x^i}e_i $$
This is a Tensor notation, as you have rightly surmised.
The position vector $\boldsymbol{x}$ is a contravariant vector, or a rank $(1,0)$ tensor.
As such, we can use the Einstein summation notation to write $\boldsymbol{x} = x^i\boldsymbol{e}_i := \displaystyle\sum_i{x^i\boldsymbol{e}_i}\,$, where $x^i$ are the vector components and $\boldsymbol{e}_i$ the basis vectors of a vector space $V$, and $i$ runs over all indices (often $1$, $2$, and $3$).
Similarly, covariant vectors or covectors (rank $(0,1)$ tensors) vectors in the dual space $V^*$, are written as $\boldsymbol{\alpha} = \alpha_i\boldsymbol{\omega}^i$, where $\boldsymbol{\omega}^i$ are the corresponding basis vectors of $V^*$ such that $\boldsymbol{e}_i \boldsymbol{\omega}^j = \delta_i^j$.
This helps when performing transformations, remembering whether a component transforms covariantly or contravariantly with the basis. It also helps with algebraic manipulations, as upper and lower indices go together, and we may contract a vector and covector as $$\boldsymbol{\alpha}\boldsymbol{x} = x^i\boldsymbol{e}_ i \, \alpha_j\boldsymbol{\omega}^j = x^i\alpha_j \delta^j_i = x^i \alpha_i\,,$$ as we expected.
Thus, in your example, using this notation, we write $$\nabla\, \phi(\boldsymbol{x}) = \partial^i \, \phi(\boldsymbol{x}) \boldsymbol{e}_i =\frac{\partial\, \phi(\boldsymbol{x})}{\partial\,x^i}\boldsymbol{e}_i\,.$$
Edit: As Nicholas Todoroff pointed out, this equation assumes orthonormality, such that the metric tensor $g_{ij} = \delta_{ij}$. A more general form of the gradient vector for arbitrary metric tensor is $$\nabla\, \phi(\boldsymbol{x}) = \mathbf{g}^{-1} \partial_i \, \phi(\boldsymbol{x}) \boldsymbol{\omega}^i = g^{ij}\frac{\partial\, \phi(\boldsymbol{x})}{\partial\,x^j}\boldsymbol{e}_i\,,$$ where $\mathbf{g}^{-1}$ is the inverse metric tensor such that $\mathbf{g}^{-1} \mathbf{g} = \mathbf{I}$, i.e. $g^{ij}g_{jk} = \delta^i_k$.
Obviously in the example, it is the case that we have orthonormality, $g_{ij} = \delta_{ij}$, and so $g^{ij}\frac{\partial\, \phi(\boldsymbol{x})}{\partial\,x^j}\boldsymbol{e}_i = \frac{\partial\, \phi(\boldsymbol{x})}{\partial\,x^i}\boldsymbol{e}_i$.