Question: The function $f(x)=x^3$ has the derivative $3x^2$ which is equal to $0$ at $x=0$ but the function $f(x)=x^3$ has neither maxima nor minima at that point. How is this possible?

Answer: 
***Hint:***

Try writing the Taylor series expansion of $\ln(x)$ at $x=1$ (this involves two cases), integrate it and then compare it with the Taylor series expansion of $x\ln(x)-x$. 

PS: But to think of this tricky solution, it is very important to know the answer already (which really doesn't makes sense). Therefore, its always better to use a hammer for a nail instead of using an axe.
Just calculate $$\left(x\ln{x}-x\right)'.$$