The most of the times when we talk about the partial derivatives of a function $f:D\subset{R^2}\rightarrow{R}$ ,we ask for the domain $D$ to be an open set of $R^{2}$. In my opinion this is very restrictive so I think, in order to determine let's say the $\partial_1f(x_0,y_0)$it would suffice for the set $D$ to include an horizontal line segment containing the point $(x_0,y_0)$. For the following example, it would be: $$\lim_{t\to x_0}\frac{f(t,y_0)-f(x_0,y_0)}{t-x_0}=\partial_1f(x_0,y_0)$$
which actually means:$$\lim_{t\to x_0^+}\frac{f(t,y_0)-f(x_0,y_0)}{t-x_0}=\partial_1f(x_0,y_0)$$
The same thing, we may deduce for $\partial_2f(x_0,y_0)$ and so on. My question here is if I miss something or have something not well understood...Thanks.
Here is something else that can go wrong when we attempt to define gradients or partial derivatives over functions that are not defined over open sets:
Define $$ \mathcal{X} = \{(x,y) \in \mathbb{R}^2 : x=2y\}$$ The set $\mathcal{X}$ is not open. Define (convex) functions $f:\mathcal{X}\rightarrow\mathbb{R}$ and $g:\mathcal{X}\rightarrow\mathbb{R}$ by $$ f(x,y) = x+y \quad, \quad g(x,y) = 3y $$ We might be tempted to say that the “gradients” of $f$ and $g$ are $$ \nabla f(x,y) = [\partial f/\partial x; \partial f/\partial y] = [1; 1] \quad, \quad \nabla g(x,y) = [\partial g/\partial x; \partial g/\partial y] = [0; 3]$$ However, notice that, in fact, $f$ and $g$ are exactly the same function! So the notion of “gradient” is not necessarily clear (or unique) when the function is not defined over an open set.
On the other hand, both $[1;1]$ and $[0;3]$ are subgradients of the function (at every point of its domain $\mathcal{X}$).