$f_n$ clearly converges to $g(x)=0$ pointwise.
Indeed, $f_n(0)=0$ for all $n$. While, for $0, we have ${1\over x}\ge1 $; whence:
$ \tag{1}|f_n(x)|=| {x\over nx+1}| ={1\over n+{1\over x}}\le{1\over n+1}\ \buildrel {n\rightarrow\infty}\over{\longrightarrow} \ 0. $
The above actually shows that the convergence is uniform: we can make $f_n(x)$ small for all $x$ by taking $n$ sufficiently large.
To be formal:
Let $\epsilon>0$. Choose $N$ so that ${1\over N+1}<\epsilon$. Then if $n\ge N$, we have, using (1): $|f_n(x)-0|=|f_n(x)|\le{1\over n+1} \le{1\over N+1}<\epsilon$ for all $0< x\le1$. Also, $|f_n(0)|=0<\epsilon$. Thus, the convergence is uniform.