I need to find all normalized vectors / versors at $(1,-1)$ for the following function: $$ f(x,y)=x^3-y^3 $$
I got $$\frac{\partial f}{\partial \vec{v}} (1,-1)= \lim_{t\to0^+}\frac{(1+tv_x)^3-(tv_y-1)^3 -(1+1)}{t} \\ 3v_x-3v_y=0 \implies v_x=v_y$$ and $$v_x^2+v_y^2=1$$ thus $\vec{v}=(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$ or $\vec{v}=(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$
Is it a correct solution?
I assume by 'normalized vectors', you mean all the normal vectors to the surface $f(x,y)$ at $(1,-1)$? A normal vector to the surface given by $f(x,y)$ will be $\nabla\! f$. But we want a unit vector, so the vector we want is $\vec{n}= \dfrac{\nabla\! f}{\|\nabla\! f\|}$. We find $$ \begin{split} \nabla\! f& = \langle 3x^2, -3y^2 \rangle \bigg|_{(1,-1)}=\langle 3,-3 \rangle \\ \|\nabla\! f\|&= \sqrt{9x^4+9y^4}= 3 \sqrt{x^4+y^4} \;\bigg|_{(1,-1)}= 3\sqrt{2} \end{split} $$ But then a unit normal vector would be $$ \vec{n}= \dfrac{\nabla\! f}{\|\nabla\! f\|}= \dfrac{\langle 3,-3\rangle}{3\sqrt{2}}= \dfrac{3\langle 1,-1 \rangle}{3 \sqrt{2}}= \left\langle \dfrac{1}{\sqrt{2}}, -\dfrac{1}{\sqrt{2}} \right\rangle $$ Of course, the 'reflection' of this vector is also a possible unit normal vector, so then we must have $$ \vec{n}= \pm \left\langle \dfrac{1}{\sqrt{2}}, -\dfrac{1}{\sqrt{2}} \right\rangle $$ So you are close, I assume there is a sign error somewhere. Although, I am not quite sure of the method as I do not know what you mean by $\partial f/\partial \vec{v}$, as I do not see how $f(x,y)$ depends on $\vec{v}$ for the partial to make sense.