Let $E$ be a normed vector space and $f_n,f:E\to\mathbb C\setminus\{0\}$ be uniformly continuous with $$\min\left(\inf_{x\in\overline B_r(0)}|f_n(x)|,\inf_{x\in\overline B_r(0)}|f(x)|\right)>0\;\;\;\text{for all }r>0\tag1$$ for $n\in\mathbb N$ and $$\sup_{x\in\overline B_r(0)}|(f_n-f)(x)|\xrightarrow{n\to\infty}0\;\;\;\text{for all }r>0.\tag2$$
Given $s,t\in[0,1]$, are we able to show $$\sup_{x\in\overline B_r(0)}\left|\frac{f_n(tx)}{f_n(sx)}-\frac{f(tx)}{f(sx)}\right|\xrightarrow{n\to\infty}0\tag3$$ for all $r>0$?
Intuitively, $(3)$ should be true and actually I think that $(1)$ is not important for that. But how can we prove it rigorously?
I think that (2) is enough, together with the fact that $f_n(x), f(x)\neq 0$ for every $x$. Specifically, let us fix $r>0$. By (2), the sequence is uniformly bounded on $\overline{B}_r$, hence there exists $M_r > 0$ such that $$ |f(x)|, |f_n(x)| \leq M_r \qquad \forall x\in \overline{B}_r,\ \forall n\in\mathbb{N}. $$ Moreover, since $|f(x)|\neq 0$ for every $x$, again by (2) there exists $m_r > 0$ and $N\in \mathbb{N}$ such that $$ m_r \leq |f(x)|, |f_n(x)| \qquad\forall x\in \overline{B}_r, \ \forall n\geq N. $$ Hence, for $n\geq N$, $t,s\in [0,1]$, $x\in \overline{B}_r$, $$ \begin{split} \left|\frac{f_n(tx)}{f_n(sx)} - \frac{f(tx)}{f(sx)}\right| &\leq \frac{1}{m_r^2} \left|f_n(tx) f(sx) - f(tx) f(sx) + f(tx) f(sx) - f_n(sx) f(tx)\right| \\ & \leq \frac{M_r}{m_r^2} \left( |f_n(tx) - f(tx)| + |f_n(sx) - f(sx)| \right)\,, \end{split} $$ and the claim follows again from (2).