Applying our power rule gets us our first derivative,

simplifying a little bit,

looking for critical points,

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.

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,

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,

and take our second derivative.

Looking for inflection points,

Again, pulling out the smaller power of x, and fractional part,

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.