Find constants so that the directional derivative of $f(x,y,z) = axy^2+byz+cx^3z^2$ has maximum value $32$ in point $P$ given the direction I am asked to find $a$, $b$ and $c$ so that the directional derivative of $$f(x,y,z) = axy^2+byz+cx^3z^2$$ has maximum value of $32$ in the point $P(1,2,-1)$ and in the direction $\overrightarrow{u} = (0,0,1)$.

What I have so far is

$$
\frac{\partial f}{\partial x} = ay^2 + 3cx^2z^2\\
\frac{\partial f}{\partial y} = 2axy+bz\\
\frac{\partial f}{\partial z} = by+2cx^3z\\
\nabla f(1,2,-1) = (4a+3c,4a-b,2b-2c)$$

$$
D_{u} f(1,2,-1) = (4a+3c, 4a-b, 2b-2c) \cdot (0,0,1) = 32 \therefore b-c = 16
$$

and I'm not sure how to proceed.

Answer: 
$$a = 3\\
b = 12\\
c = -4$$
Find values of the constants $a, b, c$ 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 $32$ in the direction parallel to the $z$-axis.
First compute the gradient:
$$\nabla f = (ay^2 + 3cz^2x^2)i + (2axy + bz)j + (by + 2czx^3)k$$
and
$$\nabla f(1, 2, −1) = (4a + 3c)i + (4a − b)j + (2b − 2c)k$$
We know that the gradient vector points in the direction of the greatest rate of change. So, if $f$
attains a maximum value in a direction parallel to the $z$-axis, this means that $\nabla f(1, 2, −1)$ points
in a direction parallel to the $z$-axis (but we don’t know if it points up or down). This gives the
equations
$$4a + 3c = 0, \quad 
 4a − b = 0, \quad
 2b − 2c = 32$$
where the final equation follows from the fact that we know that the maximum value attained by $f$
is $64$, and the fact that the maximum value of the gradient vector equals the length of the gradient
vector. Solving this, we get
$$(a, b, c) = (3, 12, −4) \text{ or } (a, b, c) = (−3,12,-4)$$
Find values of the constants $a, b, c$ 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 $32$ in the direction parallel to the $z$-axis.
First compute the gradient:
$$\nabla f = (ay^2 + 3cz^2x^2)i + (2axy + bz)j + (by + 2czx^3)k$$
and
$$\nabla f(1, 2, −1) = (4a + 3c)i + (4a − b)j + (2b − 2c)k$$
We know that the gradient vector points in the direction of the greatest rate of change. So, if $f$
attains a maximum value in a direction parallel to the $z$-axis, this means that $\nabla f(1, 2, −1)$ points
in a direction parallel to the $z$-axis (but we don’t know if it points up or down). This gives the
equations
$$4a + 3c = 0, \quad 
 4a − b = 0, \quad
 2b − 2c = 32$$
where the final equation follows from the fact that we know that the maximum value attained by $f$
is $64$, and the fact that the maximum value of the gradient vector equals the length of the gradient
vector. Solving this, we get
$$(a, b, c) = (3, 12, −4) \text{ or } (a, b, c) = (−3,12,-4)$$