$f(x,y)=x^3+3xy^2-2y^3$. Find all unit vectors, if any, such that $f_u(0,1)=\frac{6}{5}$

Question

I think that I understand what the question wants me to do:

$f(x,y)=x^3+3xy^2-2y^3$. Find all unit vectors, if any, such that $f_u(0,1)=\frac{6}{5}$

I worked out the partial derivatives:

$f_x(0,1)=3$

$f_y(0,1)=-6$

therefore $({^3_6})({^{u_1}_{u_2}})=\frac65$

but then I don't know how to work out $u_1$ and $u_2$?

Thank you!

Answer 1

$$3a-6b=\frac65$$

$$a^2+b^2=1$$

$$b=\pm \sqrt{1-a^2}$$

Solve the equation

$$(a,b)=(24/25,7/25)\ or\ (-4/5,-3/5)$$

Answer 2

By the definition of the directional derivative $f_u=\nabla f \cdot \vec{u}$ i.e. the inner product of the vector $\vec{u}=u_1\hat{x}+u_2\hat{y}$ with $\nabla f$. Since you have calculated the vector $\nabla f$, the rest of the problem is geometric. You can solve it by using algebra, i.e. solve the algebraic system $$ \begin{cases} \vec{u}\cdot\nabla f=6/5\\ u_1^2+u_2^2=1 \end{cases} $$ Can you take it from here?