Applying our power rule gets us our first derivative,
![\rm f'(x)=6\frac13x^{-2/3}+3\cdot\frac43x^{1/3}](https://tex.z-dn.net/?f=%5Crm%20f%27%28x%29%3D6%5Cfrac13x%5E%7B-2%2F3%7D%2B3%5Ccdot%5Cfrac43x%5E%7B1%2F3%7D)
simplifying a little bit,
![\rm f'(x)=2x^{-2/3}+4x^{1/3}](https://tex.z-dn.net/?f=%5Crm%20f%27%28x%29%3D2x%5E%7B-2%2F3%7D%2B4x%5E%7B1%2F3%7D)
looking for critical points,
![\rm 0=2x^{-2/3}+4x^{1/3}](https://tex.z-dn.net/?f=%5Crm%200%3D2x%5E%7B-2%2F3%7D%2B4x%5E%7B1%2F3%7D)
We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.
![0=2x^{-2/3}\left(1+2x\right)](https://tex.z-dn.net/?f=0%3D2x%5E%7B-2%2F3%7D%5Cleft%281%2B2x%5Cright%29)
When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.
Then apply your Zero-Factor Property,
![\rm 0=2x^{-2/3}\qquad\qquad\qquad 0=(1+2x)](https://tex.z-dn.net/?f=%5Crm%200%3D2x%5E%7B-2%2F3%7D%5Cqquad%5Cqquad%5Cqquad%200%3D%281%2B2x%29)
and solve for x in each case to find your critical points.
Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.
Let's come back to this,
![\rm f'(x)=2x^{-2/3}+4x^{1/3}](https://tex.z-dn.net/?f=%5Crm%20f%27%28x%29%3D2x%5E%7B-2%2F3%7D%2B4x%5E%7B1%2F3%7D)
and take our second derivative.
![\rm f''(x)=-\frac43x^{-5/3}+\frac43x^{-2/3}](https://tex.z-dn.net/?f=%5Crm%20f%27%27%28x%29%3D-%5Cfrac43x%5E%7B-5%2F3%7D%2B%5Cfrac43x%5E%7B-2%2F3%7D)
Looking for inflection points,
![\rm 0=-\frac43x^{-5/3}+\frac43x^{-2/3}](https://tex.z-dn.net/?f=%5Crm%200%3D-%5Cfrac43x%5E%7B-5%2F3%7D%2B%5Cfrac43x%5E%7B-2%2F3%7D)
Again, pulling out the smaller power of x, and fractional part,
![\rm 0=-\frac43x^{-5/3}\left(1-x\right)](https://tex.z-dn.net/?f=%5Crm%200%3D-%5Cfrac43x%5E%7B-5%2F3%7D%5Cleft%281-x%5Cright%29)
And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.
Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.