Why can't we define "the" derivative when mapping from $\mathbb{R}^2 \to \mathbb{R}$? When we map from $\mathbb{R} \to \mathbb{R}$, the derivative is given by $\lim_{x  \to a} \dfrac {f(x) - f(a)}{x-a}.$ When we map from $\mathbb{C} \to \mathbb{C}$ (basically $\mathbb{R}^2 \to \mathbb{R}^2$) we also have a definition for the derivative, $\lim_{z  \to w} \dfrac {f(z) - f(w)}{z-w}.$
However, when we map  $\mathbb{R}^2 \to \mathbb{R}$, we have only partial derivatives, $\dfrac {\partial f}{\partial x}$, $\dfrac {\partial f}{\partial y}$, and directional derivatives. Why is it that we can't define the derivative just as we do in the other $2$ cases? Is it because there is no nice way to define division between a member of $\mathbb{R}$ and a member if $\mathbb{R}^2$?

One way to look at is that, as you say, there's no way to "divide" vectors in $\mathbf{R}^2$ that really makes sense.
But you can think of it this way. When you differentiate a function $y = f(x)$ at a point $x_0$, what you're doing is saying that that function can be approximated by a linear function for values of $x$ close to $x_0$, namely
$$y \approx f(x_0) + f'(x_0)(x-x_0).$$
When you differentiate a function $z = f(x,y)$ of two variables, you likewise want to write that the function can be approximated by a linear function near $(x_0,y_0)$:
$$z \approx k + a(x-x_0) + b(y-y_0),$$
whose graph is a plane. In this approximation, the constant $k$ is $f(x_0,y_0)$, and the coefficients $a$ and $b$ are $\frac{\partial f}{\partial x} (x_0,y_0)$ and $\frac{\partial f}{\partial y} (x_0,y_0)$, respectively. In any case you need two coefficients.
