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: $$a = 3\\ b = 12\\ c = -4$$