Is this proof of uniform convergence correct?

Question

So I have that $f_n(x)=\frac{x}{1+nx^3}, n=1,2,3.. \quad$ and the question is to calculate $f'_n$, find the max and min value of $f_n$ on $R$ and show that $(f_n)$ converges uniformly on $R$. Here is how I have done:
$f'_n(x)=\frac{1-2nx^3}{(1+nx^3)^2}$ and we get that $f'_n(x)=0$ for $x=\frac{1}{\sqrt[3]{2n}}$ and this is a local maximum since we have that $lim_{x^- \to \frac{1}{\sqrt[3]{-n}}}f_n = \infty$ and vice versa $lim_{x^+ \to \frac{1}{\sqrt[3]{-n}}}f_n = -\infty \quad$
With $x=\frac{1}{\sqrt[3]{2n}}$ I get that $f_n=\frac{2}{3\sqrt[3]{2n}}$
With this information I can show that
$|f_n(x)-f_m(x)|\le max\{\frac{2}{3\sqrt[3]{2n}}, \frac{2}{3\sqrt[3]{2m}} \} \quad$ and so $\forall \epsilon >0$, we can choose $N$ s.t. $\frac{1}{N} < \epsilon$ and $\forall n,m \ge N$ we get:
$|f_n(x)-f_m(x)| < \epsilon\quad \forall x\in R$ which shows that $f_n$ is uniformly convergent on $R$
Are my calculations correct? I am more unsure about my proof of uniform convergence. Can someboy help with that?