If $f_j$ converges uniformly to $f$, does $\frac{1}{f_j}$ converge uniformly to $\frac{1}{f}$?
I think it does not have to:
Assume $f_j$ converges uniformly to $f$. In order for $\frac{1}{f_j}$ to converge uniformly to $\frac{1}{f}$, for all $\epsilon > 0$ there must exist an $N \in \mathbb{N} \ni |\frac{1}{f_j} - \frac{1}{f}| < \epsilon$. Thus, $\frac{|f - f_j|}{|f_jf|} < \epsilon$. Since $f_j$ converges uniformly to $f$, we can divide by the numerator, giving: $\frac{1}{|f_jf|} < 1$. Therefore, uniform convergence of the reciprocals is only true if $\frac{1}{|f_jf|} < 1$.
Perhaps the next question to answer, to make this proof complete, would be: If $f_j$ converges uniformly to $f$, does the reciprocal of their product have to be less than 1? (I assume: no). If I answered this question, would this be a complete proof?
... someting does not feel satisfying about this.
There are a few details which must be added, in order for the question to make sense. Firstly, we need to assume that $f$ does not vanish anywhere. Secondly, we need to assume that the domain of $f$ is compact. This way, uniform convergence of $(f_n)$ guarantees that for a sufficiently large $n$, the function $f_n$ does not vanish anywhere either.
Adding the assumptions of the previous paragraph, we show that $(1/f_n)$ does converge uniformly to $1/f$. We may assume that all functions are defined on the domain $I=[0,1]$. Denote by $m$ the minimal value of $|f|$. For a sufficiently large $n$ we have $$|f_n(x)-f(x)|<m/2$$for $x\in[0,1]$, which implies that $|f_n(x)|>m/2$ for $x\in[0,1]$. The claim now follows from the fact that the function $$y\mapsto\frac{1}{y}$$is uniformly continuous on $\{y\in\mathbb{R}|\;|y|>m/2\}$.
Remark: Instead of requiring $f$ to have a compact domain, we could add the assumption that (for a sufficiently large $n$) the functions $f_n$ are nonvanishing. In this case the series $(1/f_n)$ does not necessarily converge uniformly to $1/f$.