Answer
The unit vector ${\bf{e}}$ at $P = \left( {0,0,1} \right)$ pointing in the direction along which $f\left( {x,y,z} \right) = xz + {{\rm{e}}^{ - {x^2} + y}}$ increases most rapidly is ${\bf{e}} = \left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }},0} \right)$.
Work Step by Step
We have $f\left( {x,y,z} \right) = xz + {{\rm{e}}^{ - {x^2} + y}}$ and the point $P = \left( {0,0,1} \right)$.
By Theorem 4 of Section 15.5, $\nabla {f_P}$ points in the direction of maximum rate of increase of $f$ at $P$. So, we compute the gradient of $f$:
$\nabla f = \left( {z - 2x{{\rm{e}}^{ - {x^2} + y}},{{\rm{e}}^{ - {x^2} + y}},x} \right)$
At $P = \left( {0,0,1} \right)$ we have $\nabla {f_P} = \left( {1,1,0} \right)$.
Thus, the unit vector of $\nabla {f_P}$ is
${\bf{e}} = \frac{{\left( {1,1,0} \right)}}{{\sqrt {\left( {1,1,0} \right)\cdot\left( {1,1,0} \right)} }} = \left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }},0} \right)$
Hence, the unit vector ${\bf{e}}$ at $P = \left( {0,0,1} \right)$ pointing in the direction along which $f\left( {x,y,z} \right) = xz + {{\rm{e}}^{ - {x^2} + y}}$ increases most rapidly is ${\bf{e}} = \left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }},0} \right)$.
