I wonder if this funtion is differentiable at point $A$. I think that it is continuous so therefore I cannot use this to prove that it isn't differentiable. Can someone help me?
Edit: My main issue is that I've never worked with derivatives of parametric functions so I am not sure how to manage their derivatives. Do I have to write them as a matrix?
Compare for example to Conditions for a smooth parametric curve
First, let us check whether the function $f : (1, \infty) \rightarrow \mathbb{R}^2$ defined as $$f(t) = \begin{pmatrix} \frac{2t^3}{t^2-1} & \frac{2t^3}{(t^2+1)^2}\end{pmatrix}$$ is actually continuously differentiable.
Differentiating as usual w.r.t. $t$ gives the Jacobian $$J_f(t) = \begin{pmatrix}\frac{2t^4-6t^2}{(t^2-1)^2} & \frac{-2t^4+6t^2}{(t^2+1)^3}\end{pmatrix}.$$ $f$ is continously differentiable wherever $J_f$ is continuous. The only point where something bad can happen is clearly when $t^2-1 = 0$, i.e. $t = \pm1$. Since we assume $t>1$, $f$ is continuously differentiable (and in fact smooth) on its domain of definition.
Now, for the curve itself to be smooth (not just the function $f$), you also need to have $J_f(t) \neq 0$ for every $t\in(1,\infty)$, as explained for example here. Both coordinates of $J_f(t)$ are simulatneously zero if $t=0$ or $t = \pm\sqrt{3}$ as you can check. Since $\sqrt{3}>1$, there is indeed a $t>1$ for which the curve is not smooth. The exact point on which smoothness fails is then given by plugging $t = \sqrt{3}$ into $f(t)$: $$A = \begin{pmatrix} \sqrt{27} & \frac{\sqrt{27}}{8}\end{pmatrix}$$