Originally Posted by

**TheEmptySet** The directional derivative in the direction u is given by

$\displaystyle D_u(f)=\nabla t \cdot \frac{\mathbf{u}}{||\mathbf{u}||}$ using this gives

$\displaystyle \nabla f(1,2,6) \cdot (\mathbf{i}+2\mathbf{j}+2\mathbf{k})=9 =f_{x}(1,2,6)+2f_{y}(1,2,6)+2f_{z}(1,2,6) $

This tells us that $\displaystyle f_{z}(1,2,6)=\frac{1}{2}\left[ 9-f_{x}(1,2,6)-2f_{y}(1,2,6) \right]$

Now we have that $\displaystyle f(x,y,z(x,y))=f(x,y,2x^2+y^2) \implies z(x,y)=2x^2+y^2$

$\displaystyle \nabla f=f_x\mathbf{i}+f_y\mathbf{j}+\left(\frac{\partial f}{\partial z}\frac{\partial z}{\partial x}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial y } \right)\mathbf{k}$

$\displaystyle \nabla f=f_x\mathbf{i}+f_y\mathbf{j}+\left( 4xf_z(x,y,z)+2yf_z(x,y,z) \right)\mathbf{k}$

but we know what $\displaystyle f(x,y)$ is and can take the needed partial derivatives to finish.