Suppose $\lim_{x \rightarrow \infty} f(x) = \infty$ and $df(x)/dy = 0$ for some function $f: R \rightarrow R$.
Then, I think we have $\lim_{x \rightarrow \infty} df(x)/dy = \lim_{x \rightarrow \infty} 0 = 0$.
Can we have $d/dy(\lim_{x \rightarrow \infty} f(x)) = 0$ as well? Is the expression $d/dy(\lim_{x \rightarrow \infty} f(x))$ defined?
When we take the derivative we look over a limit: $d/dy (g(y))=\lim_{h\to0}\frac{g(y+h)-g(y)}h$.
If we set $g(y)=f(x)$ it means that $f(x)$ is a constant in respect to $y$, so we get $d/dy (g(y))=\lim_{h\to0}\frac{g(y+h)-g(y)}h=\lim_{h\to0}\frac{f(x)-f(x)}h=d/dy (f(x))=0$.
But saying $\lim_{x\to\infty}f(x)=\infty$ doesn't make a lot of sense, so I'll assume that you mean that $f(x)$ disverges to infinity.
If we add the limit of $x$ we get $\lim_{h\to0}\frac{\lim_{x\to\infty}f(x)-\lim_{x\to\infty}f(x)}h$, the next part depends in how you define that a function exists: if we say that a function that disverges to infinity exists at the limit then we can combine the limits and get $d/dy(\lim_{x\to\infty}f(x))=\lim_{h\to0}\frac{\lim_{x\to\infty}(f(x)-f(x))}h=0$,
But the majority of the times we say that disverges to infinity means that the limits doesn't exists (I saw the first case only once) so we can't combine the limits hence $d/dy(\lim_{x\to\infty}f(x))$ is not defined.
Explanation why we say that disverges to infinity is a limit that does not exist.
The simplest explanation is: we don't know the rate for which the limit changes.
Let me give you an example: say we have the function $f(x)=x$.
Now let's take 2 limits: $\lim_{x\to x_1\to\infty}x$ and $\lim_{x\to x_0\to\infty}x$.
If we say that the limit exists then both limits has to be the same so $\lim_{x\to x_1\to\infty}x-\lim_{x\to x_0\to\infty}x=\lim_{x\to x_1\to\infty}x-\lim_{x\to x_1\to\infty}x=\lim_{x\to x_1\to\infty}(x-x)=0$, regardless to how 'fast' they goes to infinity, so if I say that $x_1$ goes two times 'faster' to infinity, i.e. $x_1=2x_0$, then $\lim_{x\to x_1\to\infty}x-\lim_{x\to x_0\to\infty}x=\lim_{x\to 2x_0\to\infty}x-\lim_{x\to x_0\to\infty}x=\lim_{x\to x_0\to\infty}2x-\lim_{x\to x_0\to\infty}x$ because we said that the limit exists let's combine the limits: $\lim_{x\to x_0\to\infty}2x-\lim_{x\to x_0\to\infty}x=\lim_{x\to x_0\to\infty}(2x-x)=\lim_{x\to x_0\to\infty}x\ne 0$
but wait, in both cases I have a limit to infinity of the function $x$ no? So I have 2 different values to the limit to infinity of the same function.