Given the function: $$ f(x,y,z) = \frac{1}{x+y+z} $$ What's the directional derivative in the direction of the gradient at $(x,y,z) =(1,1,1)$?
I calculated that: $$ \overrightarrow{\triangledown} f \;\biggr\rvert_{(1,1,1)} = -\frac{1}{9}(1,1,1) $$
Thus the directional derivative in the direction of the gradient at $(1,1,1)$ is: $$ \left(\overrightarrow{\triangledown} f \;\biggr\rvert_{(1,1,1)}\right) \cdot \widehat{\left(\overrightarrow{\triangledown} f \;\biggr\rvert_{(1,1,1)}\right)} = -\frac{1}{9}(1,1,1) \cdot \frac{1}{3\sqrt{3}}(1,1,1) = - 3\sqrt{3} $$
But that answer was marked as false. Why?
It's probably worth while doing a general case first. The directional derivative of $f$ at $P$ in the direction $\bf u$ is $$\nabla f(P)\cdot\widehat{\bf u}\ .$$ For the question you asked we have ${\bf u}=\nabla f(P)$ so this becomes $$\nabla f(P)\cdot\widehat{\nabla f(P)}\ .$$ But for any vector ${\bf v}$ we have $${\bf v}=\|{\bf v}\|\widehat{\bf v}\quad\hbox{so}\quad {\bf v}\cdot\widehat{\bf v} =\|{\bf v}\|(\widehat{\bf v}\cdot\widehat{\bf v})=\|{\bf v}\|$$ (note that this is always positive, as mentioned in the comment by Jesus RS) and the final answer, using your calculation, is just $$\|\nabla f(P)\|=\frac1{3\sqrt3}\ .$$