It is clear that if $x=0$ or $x=1$, then $f_n(x)=0$ for all $n$, so $\lim\limits_{n\to\infty}f_n(x) = 0$.
If $0\lt x\lt 1$, then $0\lt 1-x^2 \lt 1$. So the question amounts to asking why/whether $\lim\limits_{n\to\infty}nt^n=0$ if $0\lt t\lt 1$.
Note that if $0\lt t\lt 1$, then $\sum_{n=0}^{\infty}t^n$ converges to $\frac{1}{1-t}$. In fact, the sequence is the Taylor series for $\frac{1}{1-t}$ around $0$ on $(-1,1)$.
Since $\sum t^n$ is the Taylor series around $0$, within the interval of convergence we can differentiate term by term and obtain the Taylor series of the derivative. That is, $\frac{1}{(1-t)^2} = \frac{d}{dt}\;\frac{1}{1-t} = \frac{d}{dt}\sum_{n=0}^{\infty}t^n = \sum_{n=0}^{\infty}\frac{d}{dt}t^n = \sum_{n=0}^{\infty}nt^{n-1} = \sum_{n=1}^{\infty}nt^{n-1}.$ Since this series converges for any $t$ in $(-1,1)$, it follows that the sequence of terms coverges to $0$; that is, $\lim_{n\to\infty} nt^{n-1} = 0\quad\text{for all }t\text{ with }|t|\lt 1.$ Now notice that for $0\lt t\lt 1$ and $n\geq 1$, we have $0 \leq (n-1)t^{n-1} \leq nt^{n-1}.$ Applying the Squeeze Theorem, since both the constant sequence $0$ and the sequence $nt^{n-1}$ converge to $0$, it follows that the sequence $(n-1)t^{n-1}$ also converges to $0$ as $n\to\infty$. That is, $\lim_{n\to\infty} (n-1)t^{n-1} = \lim_{n\to\infty}nt^n = 0\quad\text{if }0\lt t\lt 1.$
Now putting it all together we have $\lim_{n\to\infty} nx(1-x^2)^n = x\lim_{n\to\infty}n(1-x^2)^n = 0\quad\text{if }0\lt x\lt 1,$ and the cases $x=0$ and $x=1$ are already taken care of.