= Solution
The original PDF specifies evaluation at $(1,1,1)$; the local TeX incorrectly gives $(1,1,0)$. Write the scalar field explicitly as $f=c/(x^2+y^2+z^2)$. Its <gradient> is
$$
\nabla f=-\frac{2c(x,y,z)}{(x^2+y^2+z^2)^2},\qquad\nabla f(1,1,1)=-\frac{2c}{9}(1,1,1).
$$
A <directional derivative> measures rate per unit distance, so normalize the direction to the <unit vector> $\mathbf n=(1,1,0)/\sqrt2$. Then
$$
\boxed{D_{\mathbf n}f(1,1,1)=\nabla f(1,1,1)\cdot\mathbf n=-\frac{2\sqrt2\,c}{9}.}
$$
Using the unnormalized direction vector would instead differentiate with respect to a path parameter, not distance, and would give a different scale.
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