Inverse derivative of a function Problem
We have function defined as:
$$ f(x)=x^2\ln(x)$$
What is
$$ (f^{-1})'(y_0) $$ on specific point which is $y_0=e^2$
Attempt to solve
A derivative of a inverse function would be defined as:
$$ (f^-1)'(y_0)=\frac{1}{f'(f^{-1}(y_0))} $$
We can easily define $f'(x)$
$$ f'(x)=x+2x\ln(x) $$
now the problem is we don't know what is $f^{-1}$. It seems it's not possible to compute derivative of inverse function without first computing the inverse of the function. Now i don't know to to solve inverse function of $f(x)=x^2\ln(x)$ but i can see for example what wolframalpha would suggest.
wolframalpha suggests that inverse function would be something like this:
$$f^{-1}=\pm \frac{\sqrt{2}\sqrt{x}}{\sqrt{W(2x)}}$$
where W is Lambert W function
Now combining these solutions.
$$ (f^{-1})'(y_0)=\frac{1}{(\pm \frac{\sqrt{2}\sqrt{y_0}}{\sqrt{W(2y_0)}})+2 (\pm \frac{\sqrt{2}\sqrt{y_0}}{\sqrt{W(2y_0)}})\ln(\pm \frac{\sqrt{2}\sqrt{y_0}}{\sqrt{W(2y_0)}})} $$
Now the problem is i don't know how to deal with the Lambert-W function or so called "omega function". It seems i need some help solving this problem.
It would be highly appreciated if someone could hint me in right direction. At this point i am pretty convinced this is not the best way to solve this problem. 
 A: You are looking for
$$f^{-1}(e^2) = k$$
By definition
$$f(k) = e^2 = k^2\ln k$$
And this is quite easy to see that $k=e$
From here,
$$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$$
$$(f^{-1})'(e^2) = \frac{1}{f'(e)} = \frac{1}{e + 2e\ln e} = \frac{1}{3e}$$
A: As @John Lou points out, these problems, by design, are expected to be done by inspection - the point having been carefully selected so that everything works out.
But what would happen if this was not so? For example, how would one find the value of $(f^{-1})' (x)$ at say $x = 1/2$?
As you correctly suggest, the inverse function in terms of the Lambert W function would need to be found. 
First note that as $x \to 0^+$, $f(x) \to 0$, as $x \to \infty$, $f(x) \to \infty$, and since $f$ has a single (global) minimum at
$$\left (\frac{1}{\sqrt{e}}, -\frac{1}{2e} \right )$$ 
the inverse of $f$ will have two (real) branches, there being two branches when $-\frac{1}{2e} \leqslant x < 0$ and one when $x \geqslant 0$. 
To find the inverse of $f$ we set
$$x = y^2 \ln y,$$
and solve for $y$. Doing so leads to
$$y^2 = \frac{2x}{\text{W}_\nu (2x)}.$$
Here $\nu$ denotes the two real branches ($\nu = 0$ corresponding to the principal branch; $\nu = -1$ corresponding to the secondary real branch) for the Lambert W function. By making use of the defining equation for the Lambert W function, namely
$$\text{W} (x) e^{\text{W}(x)} = x,$$
we can write
$$\frac{2x}{\text{W} (2x)} = e^{\text{W}(2x)},$$
so that
$$y^2 = \exp \left (\text{W}(2x) \right ).$$
Thus
$$f^{-1} (x) = \begin{cases}
\exp \left (\frac{1}{2} \text{W}_0 (2x) \right ), & x \geqslant -\dfrac{1}{2e}\\[2ex]
\exp \left (\frac{1}{2} \text{W}_{-1} (2x) \right ), & -\dfrac{1}{2e} \leqslant x < 0.
\end{cases}$$
Now finding its derivative, noting that
$$\frac{d}{dx} \text{W}(x) = \frac{1}{x + e^{\text{W} (x)}},$$
on differentiating we have
$$(f^{-1})' (x) = \frac{\exp \left [\frac{1}{2} \text{W}_\nu (2x) \right ]}{2x + \exp \left [\text{W}_\nu (2x) \right ]},$$
where $\nu = -1,0$. 
For $x = e^2$ the principal branch is selected and we  have
$$(f^{-1})' (e^2) = \frac{\exp \left [\frac{1}{2} \text{W}_0 (2e^2) \right ]}{2e^2 + \exp \left [\text{W}_0 (2e^2) \right ]}.$$
As $\text{W} (x)$ is the inverse of the function $x e^x$ we have the following simplification rule for the Lambert W function of
$$\text{W}_0 (x e^x) = x, \quad x \geqslant -1.$$
Thus $\text{W}_0 (2e^2) = 2,$ and we have
$$(f^{-1})' (e^2) = \frac{e}{2e^2 + e^2} = \frac{1}{3e},$$
as expected.
A far more interesting exercise however is to now find the value of the derivative of the inverse function at $x = 1/2$, which, as I pointed out at the beginning, is an example that can no longer be done using inspection.
Here we have
$$(f^{-1})' \left (\frac{1}{2} \right ) = \frac{\exp [\text{W}_0 (1)/2]}{1 + \exp [\text{W}_0 (1)]}.$$
The value of the Lambert W function when its argument is equal to one corresponds to the so-called omega constant, that is, $\text{W}_0 (1) = \Omega$. Thus
$$(f^{-1})' \left (\frac{1}{2} \right ) = \frac{e^{\Omega/2}}{1 + e^\Omega} = \frac{1}{e^{\Omega/2} + e^{-\Omega/2}} = \frac{1}{2 \cosh \left (\frac{\Omega}{2} \right )} = \frac{1}{2} \text{sech} \left (\frac{\Omega}{2} \right ).$$
