Does every positive ray intersect a deformed simplex? A topological conjecture.
This can be proved using Brouwer's fixed point theorem. Note that the case that $f(x) = 0$ for some x is trivial, so by the projection $x \mapsto x/\|x\|_1$ we can reduce the problem to whether a mapping of a simplex into itself such that every face is mapped into itself must be surjective. This is indeed the case. Suppose it were not, then there would be some simplex $S$ and a point $x\in S^\circ$ that is not in the image of $f$. There would then be a retraction $r: S\setminus\{x\} \to \partial S$. We can also find a linear mapping $m:S\to S$ that cyclically permutes all the vertices. The composition $m \circ r \circ f$ would then be a continuous mapping of $S$ into itself without fixed points.
To make this a little more explicit for the present problem, I will assume that $y_i \ne 0$ for all $i$. When this is not the case, the problem can easily be reduced to a lower dimension. For a given $f$, supposing $f(x)$ and $y$ are never dependent, we can then define a function $g:\Delta^n\to\mathbb R_+^{n+1}$ by $$ g(x)_i = f(x)_i / y_i - \min_j f(x)_j / y_j $$ This is clearly a continuous function and $g(x)_i = 0$ when $f(x)_i = 0$. Furthermore, there is always an $i$ for which $f(x)_i/y_i$ is minimal, so $g(x)_i = 0$, but $g$ does not vanish because of the independence of $y$ and $f(x)$. Hence we can define $h:\Delta^n\to\Delta^n$ by $$ h(x)_i = \cases{ g(x)_{n+1}/\|g(x)\|_1 & if $i = 1$ \\ g(x)_{i-1}/\|g(x)\|_1 & otherwise. } $$ Because $g(x)_i = 0$ for some, but not all $i$, there is for every $x$ an $i$ such that $h(x)_i = 0$, but $g(x)_i \ne 0$ and therefore $x_i \ne 0$. It follows that $h(x) \ne x$ for all $x \in \Delta^n$. This contradicts BFPT, so our assumption that $y$ and $f(x)$ are always independent must fail.