What about this method to compute $\dfrac{dx}{dy}$ if $x=y+k$? I'm trying to compute $\dfrac{dx}{dy}$ if $x=y+k$.
I tried this:
$$
\begin{align}
x&=y+k\tag{1}\\[6pt]
dx&=d(y+k)\tag{2}\\[6pt]
\frac{dx}{dy}&=\frac{d(y+k)}{dy}\\[6pt]
&=\frac{dy}{dy}+\frac{dk}{dy}\\[6pt]
&=1+0\\[6pt]
\frac{dx}{dy}&=1
\end{align}
$$
But what about step $(2)$? Is that correct? what does that really mean? Should I just jump into the next step where I just take the derivative of both expressions? Can I take $d$ like so in all the situations?
From my intuition, with $(1)$ he's saying that the value of $x$ and $y+k$ is the same, so if $x$ moves a tiny bit then $y+k$ will move that same bit, and that's where $(2)$ comes from. Does this mean that $dx$ is a little proportion of $x$, same as $y+k$'s?
 A: We don't know what step $2$ means, as far as elementary calculus is concerned. But I encourage your thought process, so do not be completely disheartened!
I think as far as ordinary differentiation is concerned, one cannot talk individually about the operators $dx$ and $dy$, but rather the compound symbol $\frac{dx}{dy}$, which is the derivative of $x$ with  respect to $y$. 
Yes, we are often taught in high school that $dx$ is an infinitesimal change in $x$ and so if two quantities $x,y+k$ are equal then their infinitesimal changes are also equal (which is step $2$), but then this is just a handy interpretation : one cannot use it to justify mathematical statements.
(Ignore if you do not understand) In integration, when we often write an expression of the form $dx = f(u)du$ while attempting integration, this is just short hand for the actual transformation given by the theorem. So the actual statement above is in no way legal as far as elementary calculus in concerned.
Hence, when the individual symbols have no meaning themselves, I think the corresponding proof does not qualify to be a legal one. Indeed, the one (and most convenient) legal proof is:
$$
\frac{dx}{dy} = \lim_{h \to 0} \frac{(y+k+h) - (y+k)}{h} = \lim_{h \to 0} \frac{h}{h} = 1
$$
(The derivative should technically be at a point, but here it's the same at all points).
One last point :

If you are doing rough work (as  compared to fair work) to calculate derivatives, then a method of this kind, while not rigorous, may be helpful in at least gaining some knowledge of the answer, and is far easier to carry out. Therefore, if you are working on rough paper with no mathematician looking over your shoulder, then you are free to make logical jumps of this nature. However, in a rigorous fashion, the above proof cannot be entertained.

A: Step 2 is a bit weird as you've suspected. $dx$ is essentially a "vanishingly small" change in $x$, but in standard analysis infinitesimal numbers don't exist. In the real numbers, anything "vanishingly small" is just $0$. In other words, $(2)$ is basically just saying $0 = 0$.
If you want to think about what it all means, I think the best bet is to use the definition below of the derivate, where $x= f(y)= y + k$.
\begin{align*}
\frac{\mathrm{d}x}{\mathrm{d}y} = f'(y) = \lim_{h\rightarrow0}\frac{f(y+h)-f(y)}{h} = \lim_{h\rightarrow0}\frac{(y+h+k)-(y+k)}{h} = \lim_{h\rightarrow0}\frac{h}{h} = 1
\end{align*}
By thinking of it this way, rather than working with $dx$ and $dy$ individually as infinitesimals, you're calculating the ratio between $\Delta x$ and $\Delta y$ as they both tend to zero (which makes more sense mathematically).
A: Traditionally $d(y+k) = dy + dk$ means an infinitely small change in $y$ plus the corresponding infinitely small change in $k$. The latter is $0$ since $k$ does not change as $y$ changes.
