For mathematics at GCE A-level, I was taught that the oblique asymptote $y=x+1$ (in graph of $y=f(x)$) becomes horizontal asymptote $y=0$ (in graph of $y=\frac{1}{f(x)}$).
However, I heard that people who do mathematics in universities learn that the oblique asymptote $y=x+1$ (in graph of $y=f(x)$) becomes non-linear asymptote $y=\frac{1}{x+1}$ (in graph of $y=\frac{1}{f(x)}$). If so, do these people draw asymptote $x=-1$ of the non-linear asymptote $y=\frac{1}{x+1}$ when asked to draw graph of $y=\frac{1}{f(x)}$ given $f(x)=x+1+\frac{1}{x}$?
I think it doesn't hurt to draw but it makes the whole graph slightly more messy.
I tried searching online for such graphs but could only find polynomial asymptotes at http://www.justinmath.com/graphing-rational-functions-with-slant-and-polynomial-asymptotes/.
Whether one asymptote "becomes" another asymptote is not a clearly defined question.
It is valid to state that if $y = f(x)$ has the asymptote $y = x+1$ (the two curves get arbitrarily close) as $x \to \infty$, then $y = \frac1{f(x)}$ has the asymptote $y = \frac1{x+1}$ as $x \to \infty$. This remains valid if we replace $x+1$ by any expression that goes to $\infty$ with $x$.
It is also valid to go further and say that since $y = \frac1{x+1}$ has the asymptote $y=0$ as $x \to \infty$, so does $y = \frac1{f(x)}$. If two things get arbitrarily close to the same thing, they get arbitrarily close to each other.
Both statements are true, though I think the second is more useful than the other, since $y=0$ is a simpler function than $y = \frac1{x+1}$. Some people might not be willing to consider the $y = \frac1{x+1}$ statement at all, if they are not interested in that kind of asymptote. They don't really lose anything by it - but that has no bearing on whether it's true or false.
But to answer your question, just because $y = \frac1{f(x)}$ has the asymptote $y = \frac1{x+1}$ as $x \to \infty$, and $y = \frac1{x+1}$ has the vertical asymptote $x=-1$ as $x \to -1$, it is not valid to conclude any relationship between $y = \frac1{f(x)}$ and $x=-1$. The limits $x \to \infty$ and $x \to -1$ are different, and asymptotes are all about behavior in a limit.