Meaning of the derivative $\frac{d f(x,y)}{dx}$ Suppose we are given a function $f(x,y) = x^2 + y$ defined over $\mathbb{R}^2$. Then $$\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dx} = 2x + \frac{dy}{dx}$$ and $$\frac{\partial f}{\partial x} = 2x$$
How do I interpret $\frac{df}{dx}$, both mathematically and intuitively?
We know derivatives refer to tangents and in the case of $\frac{\partial f}{\partial x}$, we fix the $y$ and look at the tangent of the now-single-variable function $g(x) = f(x,y_0)$. I don't know what it means in the case of $\frac{df}{dx}$.
 A: Things depend a lot on whether $y$ is ultimately a function of $x$.
If $y$ is a function of $x$...
Scenario
In this kind of case, we might have a physical situation in mind.
For example, suppose we have a conical tank that has some amount of
water in it. Then the volume of water is given by the volume formula
for all cones: $V_{1}(h,r)=\dfrac{1}{3}\pi r^{2}h$. But the shape
of the tank forces a ratio between $h$ and $r$, say $r=\dfrac{h}{2}$.
So there's another function that gives the volume of the water in
terms of just the height: $V_{2}\left(h\right)=V_{1}\left(h,\dfrac{h}{2}\right)=\dfrac{1}{3}\pi\left(\dfrac{h}{2}\right)^{2}h=\dfrac{\pi}{12}h^{3}$.
In this scenario, $\dfrac{\partial V_{1}\left(h,r\right)}{\partial h}=\dfrac{1}{3}\pi r^{2}$.
And in the case of our tank, this could be written as $\dfrac{1}{3}\pi\left(\dfrac{h}{2}\right)^{2}=\dfrac{\pi}{12}h^{2}$.
And we also have $\dfrac{\mathrm{d}V_{2}\left(h\right)}{\mathrm{d}h}=\dfrac{\pi}{4}h^{2}$.
By the chain rule, these are related as
$$
\dfrac{\mathrm{d}V_{2}\left(h\right)}{\mathrm{d}h}=\left.\left(\dfrac{\partial V_{1}\left(h,r\right)}{\partial h}\dfrac{\mathrm{d}h}{\mathrm{d}h}+\dfrac{\partial V_{1}\left(h,r\right)}{\partial r}\dfrac{\mathrm{d}r}{\mathrm{d}h}\right)\right|_{r=h/2}
$$
$$
=\left.\left(\left(\dfrac{1}{3}\pi r^{2}\right)\left(1\right)+\left(\dfrac{2}{3}\pi rh\right)\left(\dfrac{1}{2}\right)\right)\right|_{r=h/2}
$$
$$
=\left(\dfrac{1}{12}\pi h^{2}\right)\left(1\right)+\left(\dfrac{1}{3}\pi h^{2}\right)\left(\dfrac{1}{2}\right)
$$
$$
=\dfrac{\pi}{4}h^{2}\checkmark
$$
Meaning
But what does this mean?
$\dfrac{\mathrm{d}V_{2}\left(h\right)}{\mathrm{d}h}=\dfrac{\pi}{4}h^{2}$
means that as you increase the height of the water $h$ by a tiny
amount $\Delta h$, the volume of the water in the tank would increase
by about $\dfrac{\pi}{4}h^{2}\Delta h$.
And $\dfrac{\partial V_{1}\left(h,r\right)}{\partial h}=\dfrac{1}{3}\pi r^{2}$
means that if you increased the height of a cone by $\Delta h$ but
did {}not{} change the radius accordingly (so that the shape of
your new theoretical cone no longer matches that of the tank), then
the volume would increase by about $\dfrac{1}{3}\pi r^{2}\Delta h$.
The chain rule calculation above breaks down the water's volume increase
into two parts: the $\dfrac{\partial V_{1}\left(h,r\right)}{\partial h}$
part that's due to the height increase alone, and the other part of
the increase that's due to the naturally-increasing radius ( $\dfrac{\partial V_{1}\left(h,r\right)}{\partial r}\dfrac{\mathrm{d}r}{\mathrm{d}h}$).
Now, all that said, it's relatively common (especially in science?)
to denote $V_{1}$ and $V_{2}$ by the same symbol, like $V$, since
they both represent the volume. Then we might say something like "$\dfrac{\mathrm{d}V}{\mathrm{d}h}$
is the rate of the entire change in the water's volume as height changes
and $\dfrac{\partial V}{\partial h}$ is the change due to the height
change alone."
Graphically
We can think about both of these values graphically, forgetting about
the tank situation entirely.
Firstly, consider $\left.\dfrac{\partial V_{1}\left(h,r\right)}{\partial h}\right|_{\left(h,r\right)=\left(2,1\right)}$.
This means we take the graph of $V_{1}$ in 3D, slice it with the
plane $r=1$ (since the partial derivative with respect to $h$ should
keep $r$ constant), and then look at the slope of the intersection
at the point with $h=2$, to get $\dfrac{1}{3}\pi\left(1\right)^{2}=\dfrac{\pi}{3}\approx1$.
See the situation in this graph:

Now, $\dfrac{\mathrm{d}V_{2}\left(h\right)}{\mathrm{d}h}$ involves
$V_{2}\left(h\right)=V_{1}\left(h,\dfrac{h}{2}\right)$. So we should
slice the graph of $V_{1}$ with $r=\dfrac{h}{2}$. Then, because
we want the change with respect to $h$, we shouldn't look at the
slope in that plane (that would be a sort of directional derivative of $V_{1}$ with value $\dfrac{2\pi}{\sqrt{5}}\approx2.8<3$),
but rather the slope of the projection onto the $hv$-coordinate plane
(corresponding to the function $V_{2}$).
Here's the situation in 3D:

Here's a viewing angle that suggests the projection and helps us see the slope $\dfrac{\pi}{4}\left(2\right)^{2}=\pi>3$

We can compare the two values in the same picture with both planes
at once:

If $y$ is not a function of $x$...
Now, if $y$ is not considered a function of $x$ in any way (and
vice versa), then there is no function like $V_{2}$ to write and
no plane like $r=h/2$ to cut with. In a normal mathematical way of
writing things, all we have is $\mathrm{d}f=\dfrac{\partial f}{\partial x}\mathrm{d}x+\dfrac{\partial f}{\partial y}\mathrm{d}y$.
(Assuming $f$ is differentiable) this basically means that if we
increase $x$ by a tiny amount $\Delta x$ and increase $y$ by a
tiny amount $\Delta y$, then $f$ will increase by approximately
$\dfrac{\partial f}{\partial x}\Delta x+\dfrac{\partial f}{\partial y}\Delta y$,
so we might write $\Delta f\approx\dfrac{\partial f}{\partial x}\Delta x+\dfrac{\partial f}{\partial y}\Delta y$.
We could divide that equation by $\Delta x$ if it's not zero, to
get $\dfrac{\Delta f}{\Delta x}\approx\dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial y}\dfrac{\Delta y}{\Delta x}=\dfrac{\partial f}{\partial x}+\dfrac{\dfrac{\partial f}{\partial y}\Delta y}{\Delta x}$,
meaning something like "the relative increase of $f$ compared to
the increase in $x$ is approximated by the sum of (a fixed term for
how much change in $x$ alone affects $f$) and (a relative term based
on any change in $y$)".
Now, this is probably not common in this context, but I suppose we
could symbolically divide everything in the original equation by $\mathrm{d}x$
to write $\dfrac{\mathrm{d}f}{\mathrm{d}x}=\dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial y}\dfrac{\mathrm{d}y}{\mathrm{d}x}$
to represent the above approximate equation in a notation with differentials.
This is not useless, as we could then apply various dependencies of
$y$ on $x$ (e.g. various shapes of conical tank) to this equation
directly; we would then be treating the $f$ on the left like $V_{2}$
and the $f$s on the right like $V_{1}$.
A: The two derivatives you provided in your question will coincide as there is no relation between $x$ and $y$. However, if one depends on the other (or let's say that they both depend on $t$) then the situation is different.
For example, if instead you had:
$$f(t)=x(t)^2+y(t)$$
then
$$\frac{df(t)}{dt}=2x(t)\frac{dx(t)}{dt}+\frac{dy(t)}{dt}$$
