Question:
The implicit funtion $z(x,y)$ is given by the equation $\ln{(x+yz)} = xy^2z^3$. Find the directions wich the directional derivative of $z(x, y)$ on the point $(0,-1)$ has value 1.
My attempt:
Since $D_u = <f_x,f_y,f_z> \cdot \ u$ I've tried calculate $f_x , f_y$ and $f_z$ using implicit differentiation finding:
$\frac{\partial{z}}{\partial{x}} = \frac{\frac{\partial{F}}{\partial{x}}}{\frac{\partial{F}}{\partial{z}}} = z^5$
$\frac{\partial{z}}{\partial{x}} = \frac{\frac{\partial{F}}{\partial{y}}}{\frac{\partial{F}}{\partial{z}}} = -z$
But I dont't know if I can do $f_z = \frac{\partial{z}}{\partial{z}} = 1$ and calculate $D_u$ using $f(x,y,z) = <z^5, -z, 1>$
Someone know how to solve this problem? And if there is something wrong on the begin of my solution?