For $x>0$ let $f(x)=x^{2/3}(6-x)^{1/3}$ and $g(x)=x\ln(x),$ then find the number of solutions of $f(x)=g(x)$ 
For $x>0$ let $f(x)=x^{2/3}(6-x)^{1/3}$ and $g(x)=x\ln(x),$ then prove that $f(x)=g(x)$ only has one real solution

Since $g(x)$ is always positive $f(x)$ will only be defined in $(0,6)$.
$\displaystyle f'(x)=\frac{2(6-x)^{1/3}}{3x^{1/3}}+\frac{x^{2/3}}{3(6-x)^{1/3}}$, since $f'(x)$ is always positive $f(x)$ is an increasing function. $g'(x)$ is also positive which means that $g(x)$ is also an increasing function. None of the functions seems to have a maxima or a minima, so I can't figure out how to compare them.
Edit: I figured out that $g(x)$ should be a concave upward function, and $f(x)$ is a concave downward function through double differentiation but I still can't figure out how to prove this :(

You write "g(x) is always positive" but that's not true, $g(x)$ is negative for $0<x<1$.
With this observation, your concavity argument gives the solution. Consider the interval $x \in (0,6)$.
You have that
$f(0) = g(0) = 0$, further $g'(0) <0$ and  $f'(0) >0$.  So for $x \to 0^+$ we have that $g(x) < 0$  and  $f(x) > 0$, or $f>g$.
Now  $g''(x) = 1/x > 0$ and $f''(x) =  {-8/(x^{4/3} (6 - x)^{5/3})} < 0$. Since $g$ is concave and $f$ is convex, this means that, since they intersect at $x=0$, they can at most have one intersection for $x >0$. You don't have to compute the point of intersection. Simply notice that $f(x=6)=0$ and $g(x=6)>0$, so $g>f$. Since this inverts $f>g$ at $x\to 0^+$, there must be exactly one intersection.
