Find the directional derivative of $f(x,y,z)=3xy+z^2$ at the point $(5,1,−4)$ in the direction of a vector making an angle of $π/3$ with $∇f(5,1,−4)$.

$f_\vec u(5,1,−4)=D_\vec uf(5,1,−4)=?$

I know how to do directional derivative questions but I have no idea about this one. I'm guessing that I'm thinking about the question wrong.

$D_\vec uf=f_x\vec u_x+f_y\vec u_y+f_z\vec u_z =\nabla f \cdot \vec u$ where $\|\vec u\|=1$

So $\nabla f= \langle 3y, 3x, 2z \rangle$ and $\nabla f(5, 1, -4)= \langle3,15,-8\rangle$

Then it says $\vec u$ makes a $\pi/3$ angle with $\nabla f(5, 1, -4)$ which would mean

$\nabla f(5, 1, -4) \cdot \vec u = \|\nabla f(5, 1, -4)\| * \| \vec u\| * \cos(\pi/3)$ and knowing $\|\vec u\| = 1$ gives

$3\vec u_x+15\vec u_y-8\vec u_z=\sqrt {298} * \cos(\pi/3)$

and with $\vec u_x^2+\vec u_y^2+\vec u_z^2=1$ gives and unsolvable system of equations (I think)

So... i'm wondering where I went wrong.