Answer
$f'(-1)$ is not defined.
$f'(-\frac{2}{3})=0.$
Work Step by Step
Product Rule $(f’(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$
$u(x)=-3x ;u’(x)=-3 $
$v(x)=\sqrt{x+1} ;v’(x)=\dfrac{1}{2\sqrt{x+1}}. $
$f'(x)=(-3x)(\dfrac{1}{2\sqrt{x+1}})+(-3)(\sqrt{x+1})$
$=\dfrac{1}{2\sqrt{x+1}}(-3x-6(x+1))=\dfrac{-9x-6}{2\sqrt{x+1}}; x\ne-1$
$f'(-1)$ is undefined.
$f'(-\frac{2}{3})=\dfrac{-9(-\frac{2}{3})-6}{2\sqrt{(-\frac{2}{3})+1}}=0.$
