We know from chain rule:
$$\dfrac{dy_x}{dx}=\dfrac{dy_u}{du}\dfrac{du_x}{dx}$$
Then will the following also be true?
$$dy_x=\dfrac{dy_u}{du}du_x$$
Is there a proof or reasoning behind it?
In limit definition, $dy_x$ represents the quantity which $\triangle{y}$ tends to when $\triangle{x}$ tends to zero. That is, $dy_x$ is zero. Same with $du_x$. By non-standard analysis, $dy_x$ and $du_x$ represents the infinitesimal change in $y$ and $u$ respectively when there is an infinitesimal change in $dx$.
Edit
Keeping the proof aside, is the infinitesimal version of chain rule mentioned in the question true or false? It seems to me that the differential article from wikipedia says it is true.
I'm not sure what you mean by "$dy$ is zero." But certainly your equation makes sense if you interpret $dy$ as a differential form.
For a function $f:M\to N$, the differential $df_x(\Delta x)$ maps tangent vectors $\Delta x$ on $M$ at $x$ to tangent vectors $\Delta f$ on $N$ at $f(x)$; in other words, $df$ is exactly the map from infinitesimal changes in $x$ to infinitesimal changes in $f$.
From linearity of the derivative it follows that $df_x$ is a linear function of $\Delta x$; therefore $df_x$'s action can be written in coordinates as tensor multiplication $[df_x]\Delta x$.
For a composition $f = y \circ u$ the chain rule then reads $$df_x(\Delta x) = dy_{u(x)}(du_x(\Delta x))$$ or $$df_x = [dy_{u(x)}]du_x;$$ an infinitesimal change in $x$ is first piped through $du$ to give an infinitesimal change in $u$, then through $dy$ to give an infinitesimal change in $f= y\circ u$.