It is not "fundamentally incorrect." It is informal and intuitive. It is true that Leibniz notation does not rigorously prove anything, but often it is more important (in my opinion) to have an intuitive understanding of why things make sense than of how to prove them formally.
A derivative is a measure of how much one variable changes in response to a change in another variable. So the symbol "$dy$" is the change in y which - taking y as a dependent variable - has been caused by a change in, say, $x$. That change we will call $\Delta x$ or $dx$. The ratio of the two changes $$\frac{dy}{dx}$$ represents how much y is altered in proportion to alterations in x (for small changes).
In this light, the chain rule should be obvious. Say z depends on y, and y in turn depends on x. $$z=z(y),\hspace{2mm} y=y(x)$$ If we change x, what happends to z? Well, when y changes by $\Delta y$, z will change by $$\frac{dz}{dy}\Delta y$$ So how much has y changed? It has changed by $$\Delta y=\frac{dy}{dx}\Delta x$$ In total then, the change in z is $$\Delta z=\frac{dz}{dy}\frac{dy}{dx}\Delta x$$ The ratio of these changes is just $$\frac{\Delta z}{\Delta x}=\frac{dz}{dy}\frac{dy}{dx}$$ In other words, we find the rate of change of z by using its derivative with respect to y, and then incorporating how much y has actually been effected by the independent variable x.
We could even make things more fancy. What would $$\frac{d\big(\sin(x)\big)}{d\big(\sin(x)\big)}$$ equal (assuming that you accept it as coherent)? Clearly if $\sin(x)$ changes by some amount, the dependent function $\sin(x)$ will change by an equal amount. So the derivative is $1$. It is not a mistake that $$\frac{dx}{dx}=1$$ The changes are equivalent and so "cancel." You can also interpret this as saying the "line $y=x$ has a slope of $1$ at all points," but the Leibniz notation is designed for another interpretation.
Inverse functions pose a similar issue. You probably know that $$\frac{d}{dx}\left(f^{-1}(x)\right)=\frac{1}{f'\left(f^{-1}(x)\right)}$$ With differentials, it is easier to see why this makes sense. Clearly $$\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$$ (Remember, I don't say "clearly" because that is simple algebra - I say it because if you understand what the derivative really means, you should see why an application of said algebra is logically valid.) The change in x and the change in y, as long as the function is actually invertible, remain the same. The inverse function will have a derivative that is the reciprocal of the original function.
-----Are there "counterexamples" to this version of differentials?-----
I would argue that there are not (and bring them on if you think you've found one!). The one which is most commonly exhibited is the following interesting formula. Let $$z=f(x,y)=0$$ Then $$\frac{dy}{dx}=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$$ If we were to "cancel" the differentials, we would mistakenly deduce the false claim $$\frac{dy}{dx}=-\frac{dy}{dx}$$ where I have changed partials back to normal derivatives.
This example, however, is based on a simple misunderstanding of what the symbols represent. The "$\partial f$" which occurs in $$\frac{\partial f}{\partial x}
$$ is the change in $f$ resulting from a change in the first variable $x$. The "$\partial f$" which occurs in $$\frac{\partial f}{\partial y}
$$ is the change in $f$ resulting from a completely different change in the second variable $y$. In other words, the two "$\partial f$"s are different quantities. They cannot be canceled. I would add that anti-intuitionists make Leibniz roll over in his grave every time they think he would have fallen for such an elementary error on the basis of what I consider to be the most brilliant notation ever devised.
Moreover, a simple 'proof' of the equation follows from basic reasoning about differentials. We have that $$\Delta f=\frac{\partial f}{\partial x}\Delta x+\frac{\partial f}{\partial y}\Delta y$$ Using the fact that $$\Delta y=\frac{dy}{dx}\Delta x$$ we get $$\Delta f=\Delta x\left(\frac{\partial f}{\partial x}+\frac{dy}{dx}\frac{\partial f}{\partial y}\right)$$ Since $f(x,y)=0$ over the curve, it will not change. Hence $\Delta f=0$ and the equation follows immediately.
A derivative divides two quantities which are both very small. An integral does just the opposite: it multiplies two quantities, one of which is large (the sum), and one of which is small (the "$\Delta x$").
The integral $$\int$$ looks like an elongated "S" on purpose - it is meant to be a "sum." It is a sum of quantities multiplied by the intervals over which they are evaluated: $$\sum{f(x_i)\Delta x_i}$$ Note that the $\Delta x$ still means the same thing; it is how much we have allowed $x$ to change. So what would happen if we integrated (summed) differentials (differences)? In other words, what is $$\int{d\big(f(x)\big)}$$ Well, if we add up all the changes over the sub-intervals, we get the net (or "total") change. Algebraically, $$\sum{\Delta f(x_i)}=\big(f(x_1)-f(x_0)\big)+\big(f(x_2)-f(x_1)\big)+\cdots +\big(f(x_n)-f(x_{n-1})\big)=f(x_n)-f(x_0)$$ because the sum telescopes. In other words, we have that $$\int{d\big(f(x)\big)}=f(x)$$ and if bounds are specified, we take the change in $f$ at the bounds (that is, $f(b)-f(a)$). The more familiar form of this theorem is $$\int{f'(x)\hspace{1mm}dx}=f(b)-f(a)$$ Once again, $$d\big(f(x)\big)=f'(x)dx$$ It all clicks!
Now let me re-write your example to see if it makes more sense. We can say that $$\int{\cos(3x)dx}=\frac{1}{3}\int{\cos(3x)3dx}=\frac{1}{3}\int{\cos(3x)d(3x)}=\frac{1}{3}\sin(3x)$$ Or another "fancy" one: We could say that $$\int{2\sin(x)d\big(\sin(x)\big)}=\sin^2(x)$$ Why? Because we are adding up changes in $\sin^2(x)$ - observe that $d(\sin^2(x))=2\sin(x)d(\sin(x))$.
If you have had multi-variable calculus, we can draw the examples further. Suppose we are asked to compute a line integral of the form $$\int{xdy+ydx}$$ The "Leibniz" way to solve this problem is as follows (think about the product rule). $$\int{xdy+ydx}=\int{d(xy)}=xy$$ The "normal" way to solve it would go something like this: We check that $$N_x-M_y=0-0=0$$ so the field is conservative. We integrate $f_x=y$ to get $$f(x,y)=xy+\phi(y)$$ Then $$f_y=x+\phi'(y)=x$$ so (after some work) for $f(x,y)=xy+c,\hspace{3mm} c\in\mathbb{R}$, we have $\nabla f=(y,x)=(M,N)$ and therefore the integral is $f=xy+c$.
Or take the famous Integration By Parts formula $$\int{udv}=uv-\int{vdu}$$ Most people don't seem to have any intuition for why this is true. However, if we manipulate it slightly we get $$\int{udv}+\int{vdu}=uv$$ Does that not look suspiciously like what we had just above? Can there really be a connection between integration by parts and vector line integrals?? YES!! They are all based on the same intuitive ideas.
Don't get me wrong: it is immensely important to understand the real definition of limits, and to witness how they can be used to systematically build up calculus into a formal and rigorous structure. But this should not be done at the cost of losing an informal grasp of what calculus is really about, and how it was discovered in the first place (hint: Newton did not have an epiphany involving epsilons and deltas when he saw an apple fall). I do not have space to talk about more objects in calculus, but I believe that, with enough thought, they all turn out to be deeply related and surprisingly intuitive.
In many concrete situations, you may have some specific expression on hand. Consider the density of the normal distribution, $f(x,\mu,\sigma)=(2\pi\sigma^2)^{-1/2}\exp(-|x-\mu|^2/2\sigma^2)$. If you simply write
$$
\left((2\pi\sigma^2)^{-1/2}\exp(-|x-\mu|^2/2\sigma^2)\right)',
$$
it will of course be completely unclear what you mean. If you want, you can define $f_{\mu,\sigma}(x)=f(x,\mu,\sigma)$, and write $f_{\mu,\sigma}'$, but it quickly becomes cumbersome to define new functions every time you wish to take a derivative. It is much simpler to simply write, say
$$
\left.\frac{d}{dx}\right|_{x=1}(2\pi\sigma^2)^{-1/2}\exp(-|x-\mu|^2/2\sigma^2)
$$
in place of first defining $f_{\mu,\sigma}$ and then writing $f_{\mu,\sigma}'(1)$.
This is much the same reason that it is useful to have the $dx$ appearing somewhere when you integrate, instead of a more general $d\mu$, where $\mu$ is a measure. It is notationally less pretty, but much more flexible.
Best Answer
You should know that the differential at a point $\mathbf x_0$ of a function $\;\mathbf R^m\longrightarrow \mathbf R^n$ is the linear map $\:\ell:\mathbf R^m\longrightarrow \mathbf R^n$, that yields the best linear approximation of $f(\mathbf x_0)$ in a neighbourhood of $\mathbf x_0$, in the sense that we have $$f(\mathbf x_0+\mathbf h)=f(\mathbf x_0)+\ell(\mathbf h)+o\bigl(\|\mathbf h\|\bigr).$$ This differential is denoted $\:\mathrm d f_{\mathbf x_0}$ (or simply $\mathrm df$ for the differential at a generic point). A linear function is of course its own differential.
With the usual abuse of language that denotes a function $f$ by its value at a given point, we obtain that the differential of the $i$-th projection $\:p_i:\mathbf x=(x_1,x_2,\dots,x_m)\longmapsto x_i$ is denoted $\mathrm dx_i$.
As an example, in the case of a function of a single variable $x$, the linear map defining the differential simply corresponds to the equation of the tangent to the graph of $f$ with abscissa $x_0$: $$f(x_0+h)=f(x_0)+f'(x_0)h,\enspace\text{i.e. }\quad \mathrm df_{x_0}:h\longmapsto f'(x_0)h,$$ that we may write as $\enspace\mathrm df=f'(x)\,\mathrm dx$. This notation is generalised to functions of $m$ variables under the form $$\mathrm df=\frac{\partial f}{\partial x_1}\,\mathrm dx_1+\frac{\partial f}{\partial x_2}\,\mathrm dx_2+\dots+\frac{\partial f}{\partial x_m}\,\mathrm dx_m. $$ The usual formulæ for the derivatives have a ‘differential version’: