Part

**(c)**,

**(d)** and

**(e)** of the following problem:

http://img368.imageshack.us/img368/568/61463247.gif
My attempt (please correct me where I am wrong)

**(c)** $\displaystyle \hat{u} = \| u \| = \sqrt{1^2 + 1^2} = \sqrt{2}$

so

$\displaystyle \hat{u} = \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix}$

From part (a) I know that $\displaystyle f_x(0,0)=f_y(0,0)=0$. Therefore:

$\displaystyle \nabla f(0,0)=f_x(0,0) i + f_y(0,0)j = 0$

Therefore $\displaystyle D_u f(0,0)= \nabla f(0,0). \hat{u} = 0$

**(d)** Letting $\displaystyle y=x$ we will have $\displaystyle f(x)=\sqrt[3]{2x^3}$. Therefore

$\displaystyle f'(x)= \frac{2^{(1/3)}}{(x^3)^(2/3) x^2}$

$\displaystyle \nabla f = f(0)i = 0$

Therefore $\displaystyle D_u f(0,0)= \nabla f(0). \hat{u} = 0$