Find the inverse of $f(x)=x^3+x+1$. $$f(x)=x^3+x+1$$
I didn't learn this at school and I want to know how I can get the inverse of this function.  
Do you use differentiation?  
I have this solution but I don't understand what it means $(f^{-1})'(3)$=$1$.  
Is that differentiation?

First you have to check that $f$ is bijective (assuming that the domain and the codomain are both $\Bbb R$).  The surjectivity is clear as $f$ is continuous and $\lim_{x\to\pm\infty}f(x)=\pm\infty$.  For injectivity, we have
$$f(x)=f(y)\iff x^3-y^3+x-y=0\iff (x-y)(x^2+xy+y^2+1)=0.$$
Since $x^2+xy+y^2+1=(x+y/2)^2+3(y/2)^2+1>0$, we see that $$f(x)=f(y) \iff x=y,$$ so $f$ is injective.  Hence, $f$ is a bijection, and we can find the inverse.
Suppose that $f(x)=t$.  Then, $x^3+x=t-1$.  We assume alla Cardano's method that $x=u+v$ such that $x^2+1=u^2-uv+v^2$, so $u^3+v^3=x(x^2+1)=x^3+x=t-1$.  Now, $$u^2-uv+v^2=x^2+1=(u+v)^2+1\implies uv=-\frac13\implies u^3v^3=-\frac1{27}.$$
That is, $u^3$ and $v^3$ are roots of $z^2-(t-1)z-\frac1{27}$, which means
$$\{u^3,v^3\}=\left\{\frac{(t-1)\pm\sqrt{(t-1)^2+\frac{4}{27}}}{2}\right\}.$$
So, wlog, we can take
$$u=\sqrt[3]{\frac{(t-1)+\sqrt{(t-1)^2+\frac{4}{27}}}{2}}\wedge v=\sqrt[3]{\frac{(t-1)-\sqrt{(t-1)^2+\frac{4}{27}}}{2}}. $$
This gives
$$f^{-1}(t)=x=u+v=\sqrt[3]{\frac{(t-1)+\sqrt{(t-1)^2+\frac{4}{27}}}{2}}+\sqrt[3]{\frac{(t-1)-\sqrt{(t-1)^2+\frac{4}{27}}}{2}}.$$
