Understanding partial derivative Let's say I have function
$z=f(x,y)=x^2+xy+y^2$
Please verify if my understanding of partial derivative is correct..I have put it in my words below:
I understand that a partial derivative with respect to $x$ gives me equation of slope if $y$ is kept constant. When we keep $y$ constant, we take a plane perpendicular to $y$ axis, curve $z$ will intersect that plane and that intersection will be a line or a curve. The partial derivative with respect to $x$ is $2x+y$ and therefore that is the equation of tangent at any point on that line. That tangent line will be parallel to $x$ axis.
Now if we take partial derivative of (partial derivative of $x$) with respect to $y$ what do I get? In this case it will be derivative of $2x+y$ and it will come out to be $1$. Does it has any meaning? Also what is the total derivative in this case? what does total derivative give? In case of $z$ curve if I want to find coordinates of point that has the maximum or minimum value then how could we calculate it?
thanks
I tried wiki search but couldn't understand the concepts. I also looked at a few coursera courses but I felt that they explain calculations than explaining the concept. 
 A: Maybe it's helpful to view at the partials from a more general point.  Let $c$ be a differentiable curve defined on an open interval $I$ containing $0$ to $\mathbb R^2$ with $c(0)=p$ and $c'(0)=v$.  The standard-example for such a curve is $c(t):=p+tv$. Let $f\colon G\mapsto\mathbb R$ defined on an open subset $G$ of $\mathbb R^2$.  Let $p\in G$ and $v\in\mathbb R^2$.  We define the derivative of $f$ in $p$ at $v$, namely $d_pf(v)$ via
$$d_pf(v)=\frac{d}{dt}\Big|_{t=0}f\bigl(c(t)\bigr),$$
provided that limit exists.  Then for a fixed $p$ the map
$$ d_pf\colon\mathbb R^2\to \mathbb R\quad\text{is linear.}$$
Now we may want to find a matrix for that linear map in respect to the basis $\{(1,0)^T,(0,1)^T\}$ of $\mathbb R^2$ and $\{1\}$ of $\mathbb R$.  All we need to do is to plug in the base vectors in $d_pf$.  Well, to plug in $(1,0)^T$ we'll have to evaluate
$$\frac{d}{dt}\Big|_{t=0}f\bigl(p+t(1,0)^T)\bigr),$$
i.e., treat $y$ as constant and differentiate $f$ in respect to $x$.  We will henceforth call
$$\frac{\partial f(p)}{\partial x}:=d_pf\bigl((1,0)^T\bigr)$$
the partial derivative of $f$ in $p$.
Thus the matrix of $d_pf$ is
$$\left(\frac{\partial f(p)}{\partial x},\frac{\partial f(p)}{\partial y}\right).$$
Let's define
$$\nabla_pf:=\left(\frac{\partial f(p)}{\partial x},\frac{\partial f(p)}{\partial y}\right)^T,$$
then presence of the standard dot product we finally gain
$$d_pf(.)=\langle\nabla_pf(.),.\rangle.$$
