You are developing the right idea, but your Eq. 1 is a bit awkward. To understand why, let me say a bit about functionals and functional derivatives.
A functional acts on a function in its entirety, for example we need to know the full input function $n(r)$ in order to obtain the corresponding value for the functional:
$$\tag{1}
E[n(r)] = \int_0^1n(r)\textrm dr.
$$
The output is different for $n(r) = n^2$ versus for $n(r) = \sqrt{n},$ and each $n(r)$ function will give a different value for the functional $E$.
This is different from an ordinary composite function like $f(g(r))$ with round parentheses, because you do not need to take into account all of $g(r)$ in order to get $f(g(r))$, since $f$ is only acting on $g$ at a specific point like $r=2$, which means that $f(g(2))$ makes sense but $E[n(2)]$ is a bit awkward. Technically speaking, $n(2)$ can be considered a constant function, but a constant function of what? The integral in my Eq. 1 can still be evaluated (with respect to the variable that was already set to 2?), but the square brackets tell us that $E[n(r)]$ is meant to be evaluated based on all values of $r$, not just one of them.
Technically functionals are also functions, in which the domain contains functions (i.e. it maps functions to outputs, which can be scalars, vectors, functions, or even functionals; an example of the latter being a functional that turns polynomials into operators, like $r^2 \rightarrow \frac{\textrm{d}^2}{\textrm{d}r^2}$), but we use square brackets and round brackets to tell us how the author intends for us to use the two different types of operators.
Consider the difference between the following two derivatives:
$$\tag{2}
\frac{\textrm{d}f(g(r))}{\textrm{d}r}~~~~~~~~~~~,~~~~~~~~~~~ \frac{\delta f[g(r)]}{\delta g(r)}~~.
$$
It would be awkward to put $g(r)$ in the denominator of the first one, because the ordinary derviative $\frac{\textrm{d}}{\textrm{d}r}$ acts on functions whose domains are points (not functions), and likewise it would be awkward to put just a point like $g(r_1)$ in the denominator of the second one (otherwise why are you using the $\frac{\delta}{\delta\left(\cdot\right)}$ notation rather than the $\frac{\textrm{d}}{\textrm{d}\left( \cdot \right)}$ notation?).
Therefore, your Eq. 1 is written awkwardly.
Now consider what it really means for the functional derivative of the energy with respect to the density, to be 0. It means that changing the function $n(r)$ will not change the energy. For example if you shift the Gaussian $n(r) = e^{-r^2}$ to $n(r) = e^{-(r-0.1)^2}$, the energy does not change (if the functional derivative of $E$ with respect to $n(r)$ is 0). Contrarily, if the derivative of the energy with respect to a grid point $r_1$ is 0, it would mean:
$$\tag{3}
\left. \frac{\textrm{d}E(n(r))}{\textrm{d}r_1} = \frac{\textrm{d}E}{\textrm{d}n}\frac{\textrm{d}n}{\textrm{d}r} \right|_{r=r_1}= 0,
$$
which can happen if $n(r)$ is flat, meaning that it doesn't change if you move the grid point $r_1$.
It is not true that the density $n(r)$ has to be flat (constant with respect to $r$) in order for $\frac{\delta E}{\delta n(r)}$ to be 0. The density can take the shape of a roller coaster, and still we can have that the energy does not change when we change the shape of the roller coaster (i.e. the functional derivative of the energy with respect to the shape of the roller coaster is 0).
