Consider,
$$ f_n(x)=\frac{1-nx^2}{(1+nx^2)^2} $$
where, $$x \in \mathbb{R}, n \in \mathbb{N}$$
It is clear to me that not including $x = 0$, each function point converges to $0$ as $n \to \infty$.
The confusion is the convergence at $x = 0$. It seems to converge to both $f = 1$ and $f = \frac{-1}{8}$. On differentiating and setting equal to zero, the stationary points occur at $$x = 0, f_n(x) = 1$$ $$x_n = \pm{\sqrt{\frac{3}{n}}}, f_n(x) = \frac{-1}{8}$$ Both function values don't depend on $n$, however as $n \to \infty$,
$$x_n = \pm{\sqrt{\frac{3}{n}}} \to 0$$
Which would suggest that as $x \to 0$ and $n \to \infty$, $$f_n(x) \to \frac{-1}{8}$$ and $$f_n(x) \to 1$$
Could someone explain to me what is going on, I suspect it has something to do with converging to disjoint functions?
Does this mean the sequence of functions does not converge pointwise (and thus not uniformly)? How would I go about formally proving this with $\epsilon$ quantifers.
This is not an entirely complete answer (i.e. there's no $\epsilon,\delta$ ) but this is too long to fit in a comment. We say $f_n\to f$ pointwise if $f(x)=\lim_{n\to \infty} f_n(x)$. From this definition, it's clear that $f(0)=1$ but is zero otherwise. This is enough to prove that the convergence is not uniform, because uniform convergence would imply that $f$ is continuous.
What you have found is that there exists a sequence $x_n$ such that $\lim_{n\to \infty}x_n=0$ but $\lim_{n\to \infty}f_n(x_n)\neq f(0)$. This has no bearing on the pointwise limit because of the pointwise limit's definition. I believe that this is another proof that convergence isn't uniform, because uniform convergence should imply that those two limits are equal.