I tried to prove the uniform convergence of the sequence of functions
$$f_n \left( x \right) = \dfrac{nx}{1 + n^3x^2}$$
defined on $\left[ 0, 1 \right]$ in the following manner:-
First, it is easy to see that $f_n$ is pointwise convergent to $f \equiv 0$ since for any $x \in \left[ 0, 1 \right]$, we have $\lim\limits_{n \rightarrow \infty} \dfrac{nx}{1 + n^3 x^2} = 0$.
To see uniform convergence, first we observe that $\dfrac{nx}{1 + n^3x^2} \leq \dfrac{n}{1 + n^3x^2}$ which converges to $0$. Therefore, by snadwich lemma, $\left| f_n \left( x \right) \right|$ converges to $0$ and therefore we have the uniform convergence.
Although I did prove the uniform convergence, I think that there is a mistake in the proof. Any help will be appreciated.
Note that $f_n'(x)=\dfrac{n-n^4x^2}{(1+n^3x^2)^2}$. Therefore, $f_n$ attains its maximum at $\dfrac1{n^{3/2}}$, and that maximum is $\dfrac1{2\sqrt n}$. Since $\lim_{n\to\infty}\dfrac1{2\sqrt n}=0$, your sequence of functions converges uniformly to the null function.
Concerning your proof, it is easier to observe that $(\forall n\in\mathbb N):f_n(0)=0$ and that, if $x\in(0,1]$,$$\lim_{n\to\infty}\frac{nx}{1+n^3x^2}=0,$$since the numerator is a first degree polynomal, whereas the denominator is a cubic polynomial.