Answer
${D_{\bf{u}}}f\left( P \right) = \frac{1}{2}\sqrt 6 $
Work Step by Step
We have $f\left( {x,y,z} \right) = \sin \left( {xy + z} \right)$ and $P = \left( {0, - 1,\pi } \right)$.
The gradient of $f$ is
$\nabla f = \left( {y\cos \left( {xy + z} \right),x\cos \left( {xy + z} \right),\cos \left( {xy + z} \right)} \right)$
The gradient of $f$ at $P$ is $\nabla {f_P} = \left( {1,0, - 1} \right)$. So,
$||\nabla {f_P}|| = \sqrt {{1^2} + {0^2} + {{\left( { - 1} \right)}^2}} = \sqrt 2 $
Using Theorem 4, the rate of change of $f$ in the direction of a unit vector making an angle $\theta = 30^\circ $ with $\nabla {f_P}$ is
${D_{\bf{u}}}f\left( P \right) = ||\nabla {f_P}||\cos \theta $
${D_{\bf{u}}}f\left( P \right) = \sqrt 2 \cos 30^\circ = \frac{1}{2}\sqrt 6 $
