Here is Prob. 5, Exercises 8.14, in the book Calculus Vol. II by Tom M. Apostol, 2nd edition:
Find values of the constants $a$, $b$, and $c$ such that the directional derivative of $f(x, y, z) = axy^2 + byz + cz^2x^3$ at the point $(1, 2, -1)$ has a maximum value of $64$ in a direction parallel to the $z$-axis.
My Attempt:
We note that $$ f_x(x, y, z) = ay^2+3cz^2x^2, \qquad f_y(x, y, z) = 2axy+bz, \qquad f_z(x, y, z) = by+2czx^3. $$ So $$ f_x(1, 2, -1) = 4a+3c, \qquad f_y(1, 2, -1) = 4a-b, \qquad f_z(1 , 2, -1) = 2b-2c.$$ Thus we see that all the partial derivatives of $f$ exist and are continuous everywhere in $\mathbb{R}^3$. So $f$ is differentiable everywhere in $\mathbb{R}^3$. Therefore, the directional derivative of $f$ at any point $(x, y, z) \in \mathbb{R}^3$ and in any direction $(u, v, w)$ in $\mathbb{R}^3$ is given by $$ f^\prime( (x, y, z); (u, v, w) ) = \nabla f(x, y, z) \cdot (u, v, w) = \left( ay^2+3cz^2x^2, 2axy+bz, by+2czx^3 \right) \cdot (u, v, w) = \left( ay^2+3cz^2x^2 \right) u + \left( 2axy+bz \right) v + \left( by+2czx^3 \right) w. $$ And, in particular, the directional derivative of $f$ at the point $(1, 2, -1)$ in any direction $(u, v, w)$ is given by $$ f^\prime( (1, 2, -1); (u, v, w) ) = (4a + 3c)u + (4a-b)v + (2b-2c)w. $$ Now this derivative has its largest value when $$ (u, v, w) = \nabla f(1, 2, -1) = (4a + 3c, 4a-b, 2b-2c), $$ and this value is $$ \lVert \nabla f(1, 2, -1) \rVert^2 = (4a+3c)^2 + (4a-b)^2 + (2b-2c)^2, $$ and this direction is parallel to the $z$-axis if and only if $$ 4a + 3c = 0 \qquad \mbox{ and } \qquad 4a-b = 0, $$ which is the case if and only if $$ b = 4a \qquad \mbox{ and } \qquad c = -\frac{4a}{3}, \tag{1} $$ and so $$ \lVert \nabla f(1, 2, -1) \rVert^2 = \left( 8a + \frac{ 8a }{3} \right)^2 = \left( \frac{32a}{3} \right)^2 = \frac{ (32)^2 a^2}{9}, $$ and this value equals $64$ if and only if $$ a^2 = \frac{ 64 \times 9}{(32)^2} = \frac{9}{16}, $$ and so $$ a= \pm \frac{3}{4}. $$ Then (1) gives $$ b = \pm 3 \qquad \mbox{ and } \qquad c = \mp 1. $$
Therefore we have two triples of values for $(a, b, c)$: $$ (a, b, c) = \left( \frac{3}{4}, 3, -1 \right) \qquad \mbox{ and } \qquad (a, b, c) = \left( -\frac{3}{4}, -3, +1 \right). $$
Is my solution correct? Or, are there problems in my logic or calculations?