To show, that the inverse function is not differentiable at $\mathbf 0$. Let $f:\mathbb R^n\to\mathbb R^n$ by $f(\mathbf x)=\|\mathbf x\|^2 \mathbf x$. To show, that the inverse function is not differentiable at $\mathbf 0$. 
I am not able to find the inverse of the function. Since we know that inverse of column vectors does not exists.
 A: Think about what $f(x)$ does: it scales every vector in $n$-space by the square of its length.  In other words, it maps $x$ to a vector in the same direction, but with magnitude $\|x\|^3$ (because $x$ already has magnitude $\|x\|$).  To find the inverse, let's say $u = f(x)$.  Then we know $u$ is in the same direction as $x$, so we need only recover $\|x\|$ from $\|u\|$.  But since $\|u\| = \|x\|^3$, we have $\|x\| = \|u\|^{1/3}$.  So we can recover $x$ from $u$ by 
\begin{equation*}
f^{-1}(u) = \|u\|^{1/3} \frac{u}{\|u\|} = \|u\|^{-2/3} u
\end{equation*}
for $u \neq 0$.  We also define $f^{-1}(0) = 0$.
From here you can see that the problem with taking the derivative of $f^{-1}$ is that it will involve $\|u\|^{-4/3}$, which doesn't exist at zero.  I'll let you work the full details of the derivative out for yourself.
A: Just to clarify first: by the inverse function we mean function $g:\mathbb R^n\to\mathbb R^n$ such that $$f\circ g = g\circ f = \operatorname{id}_{\mathbb R^n}$$ which has nothing to do with "inverse of column vectors", whatever you meant by that.
This inverse function $g$ is not hard to find, as already explained by cderwin, but we don't even need to know exact formula to prove the claim. Well, we need some formula, it will be clear what I mean soon.
Here's the deal. From $f(x) = \|x\|^2 x$ we conclude that given any unit vector $e$ (i.e. $\|e\| = 1$), we have $f(te) = t^3e$. It immediately follows that for unit vector $e$ we will have $g(te) = \sqrt[3]{t\,}e$.
Now, if $g$ were differentiable at $0$, then all partial derivatives $\partial_ig_j(0)$ would exist. But,
$$\partial_ig_i(0) = \left.\frac d{dt}g_i(te_i)\right|_0 = \left.\frac d{dt}\sqrt[3]t\,\right|_0$$ doesn't exist. We conclude that $g$ is not differentiable at $0$.
