What is this limit called? Is it a different kind of derivative? (first I should notice you this is not something I can look up in a textbook, because I'm learning partial derivatives, alike I do with most Maths, as a hobby. If something below is wrong, blame the internet and please let me know)
I know I can take a mixed partial derivative of a multi-argument function, for expl.:
$$
w = f(x; y; z) = x^2 y+ y^2z + xz^2\\
\frac{\partial^2 w}{\partial x\ \partial y} = \frac{\partial w}{\partial x}\frac{\partial w}{\partial y} = \frac{\partial}{\partial y}(2xy+z^2)= 2x = ^*\frac{\partial}{\partial x}(2yz+x^2)
$$
(* = double check)
But that's taking 2 partial derivatives. What If I want more than one variable to vary at once?
$$
\lim_{h \to 0} \left( \frac{f(x+h;y+h;z)-f(x;y;z)}{h}\right)=\\
=\lim_{h \to 0} \left( \frac{(x+h)^2 (y+h)+(y+h)^2 z+(x+h)^2z^2-(x^2 y+ y^2z + xz^2)}{h}\right)=\\
=\lim_{h \to 0} \left( \frac{(x^2+2xh+h^2)(y+h)+(y^2+2hy+h^2) z+(x+h)z^2-(x^2 y+ y^2z + xz^2)}{h}\right)=\\
=\lim_{h \to 0} \left( \frac{2hxy+h^2y+hx^2+h^2x+h^3+2hyz+h^2z+hz^2}{h}\right)=\\
=\lim_{h \to 0} \left( 2xy+hy+x^2+hx+h^2+2yz+hz+z^2\right)
=2xy+x^2+2yz+z^2
$$
Note: I took a while to write this, and only now I notice that the result of the above is the sum of $\frac{\partial w}{\partial x}$ and $\frac{\partial w}{\partial y}$, while the mixed partial in the beginning is its multiplication
Nonetheless, the question remains: what is this limit called? Can it be seen as another type of derivative? What are its applications? Is there a notation in which I can write it such as the one for other derivatives?
Thank you readers in advance (and for reading, my questions are not usually short).
 A: For, what you're looking at is a special case of a directional derivative of a function from $\mathbb{R}^n$ to $\mathbb{R}$, i.e. a derivative along some direction (vector) $d$. In general, those are defined as $$
  f_d(x) = \lim_{t\to 0} \frac{f(x+td) - f(x)}{t} \text{,}
$$
where $x,d \in \mathbb{R}^n$ and $t \in \mathbb{R}$. You looked at the special case $d=(1,1,0)$, i.d. you let the first and the second argument vary simultaneously and the third not at all.
Sometimes (or often, depending on what functions you usually work with), the directional derivatives are a linear function of the direction. In that case, the function is simply called differentiable. In other words, you then have that $$
  f_{u+v}(x) = f_{u}(x) + f_{v}(x) \text{.}
$$
If that is so, then it's sufficient to know e.g. the derivatives in the direction of all the axes to compute any directional derivative. Since the derivatives in the direction of the axes are simply the ususal partial derivatives, this means that in this case $$
  f_{d} = \sum_{i=1}^n d_i \frac{\partial f}{\partial x_i}
  \quad\text{assuming that}\quad
  d = (d_1,\ldots,d_n).
$$
Which is exactly what you found - you computed $f_{(1,1,0)}$ and found it to be $1\frac{\partial f}{\partial x} + 1\frac{\partial f}{\partial y} + 0\frac{\partial f}{\partial z}$.
If this works, it's custom to arrange the partial derivatives into a vector (or matrix, if the function also produces multiple values). The derivative of a function $f \,:\, \mathbb{R}^n \to \mathbb{R}$ is thus the vector $$
  \frac{df}{dx} = f' = \left(\frac{\partial f}{\partial x_1},\ldots, \frac{\partial f}{\partial x_n}\right).
$$
Similarly to the one-dimensional case, you can then approximate $f$ around $x$ by (note that $\cdot$ is the scalar product!) $$
  f(x+d) \approx f(x) + f'(x)\cdot d \text{.}
$$
Note that this doesn't always work - one example is $$
  f(x,y) = \begin{cases}
    \frac{xy}{\sqrt{x^2 + y^2}} &\text{if $(x,y)\neq(0,0)$} \\
    0                           &\text{otherwise}
  \end{cases}
$$
at $(0,0)$. Both partial derivatives are $0$ at $(0,0)$, yet the derivative in direction $(1,1)$ isn't. This function therefore is said to be non-differentiable at $(0,0)$, even though the partial derivatives do exists. But they don't convey all the information there is to know about the function's behaviour around $(0,0)$.
